Jawaban Soal Latihan 1 Perkalian Matriks

Soal Latihan 1

1. Diketahui matriks A dan B sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}1& 2\\ -2& 3\\ 3& -1\end{array},,\right)\right)$
   
   Tentukan :
   
   1. 3A
   2. 2A
   3. 5A^t
   4. 2A^t + 5B
   5. 2B^t + 3A


2. Diketahui matriks A dan B sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}\mathrm{a}& 4\\ \mathrm{2b}& \mathrm{3c}\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}\mathrm{2c\; -\; 3b}& \mathrm{2a\; +\; 1}\\ \mathrm{a}& \mathrm{b\; +\; 7}\end{array},,\right)\right)$
   
   Tentukan nilai c jika A = 2B^t


3. Diketahui matriks A, B dan C sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$
   
   C = $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$
   
   Tentukan nilai dari :
   1. AB
   2. AC
   3. BC
   4. A^tC
   5. B^tC
   6. C^tA
   7. C^tB
   8. A^tB^t
   9. B^tA^t
   10. (AB)^t
   11. Apakah (AB)^t = B^tA^t


4. Tentukan nilai a dan b jika diketahui persamaan berikut :
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 3& -2\end{array},,\right)\right)\left(\left(,,\begin{array}{cc}6& -5\\ 2& 4\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   


5. Tentukan nilai x dan y yang memenuhi persamaan berikut :
   
   $\left(\left(,,\begin{array}{cc}1& 5\\ 4& -6\end{array},,\right)\right)\left(\left(,,\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   

Jawab

1. 1. 3A = 3 $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)=\left(\left(,,\begin{array}{ccc}-3& 9& -12\\ 6& 0& 15\end{array},,\right)\right)$
      
   
   
   2. 2A = 2 $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)=\left(\left(,,\begin{array}{ccc}-2& 6& -8\\ 4& 0& 10\end{array},,\right)\right)$
      
   
   
   3. A^t = $\left(\left(,,\begin{array}{cc}-1& 2\\ 3& 0\\ -4& 5\end{array},,\right)\right)$
      
      5A^t = 5 $\left(\left(,,\begin{array}{cc}-1& 2\\ 3& 0\\ -4& 5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-5& 10\\ 15& 0\\ -20& 25\end{array},,\right)\right)$
      
   
   
   4. 2A^t + 5B = 2 $\left(\left(,,\begin{array}{cc}-1& 2\\ 3& 0\\ -4& 5\end{array},,\right)\right)$ + 5 $\left(\left(,,\begin{array}{cc}1& 2\\ -2& 3\\ 3& -1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}-2& 4\\ 6& 0\\ -8& 10\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}5& 10\\ -10& 15\\ 15& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}3& 14\\ -4& 15\\ 7& 5\end{array},,\right)\right)$
   
   
   5. B^t = $\left(\left(,,\begin{array}{ccc}1& -2& 3\\ 2& 3& -1\end{array},,\right)\right)$
      
      2B^t + 3A = 2 $\left(\left(,,\begin{array}{ccc}1& -2& 3\\ 2& 3& -1\end{array},,\right)\right)$ + 3 $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}2& -4& 6\\ 4& 6& -2\end{array},,\right)\right)$ + $\left(\left(,,\begin{array}{ccc}-3& 9& -12\\ 6& 0& 15\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-1& 5& -6\\ 10& 6& 13\end{array},,\right)\right)$


2. B^t = $\left(\left(,,\begin{array}{cc}\mathrm{2c\; -\; 3b}& \mathrm{a}\\ \mathrm{2a\; +\; 1}& \mathrm{b\; +\; 7}\end{array},,\right)\right)$
   
   A = 2B^t
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a}& 4\\ \mathrm{2b}& \mathrm{3c}\end{array},,\right)\right)$ = 2 $\left(\left(,,\begin{array}{cc}\mathrm{2c\; -\; 3b}& \mathrm{a}\\ \mathrm{2a\; +\; 1}& \mathrm{b\; +\; 7}\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a}& 4\\ \mathrm{2b}& \mathrm{3c}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{4c\; -\; 6b}& \mathrm{2a}\\ \mathrm{4a\; +\; 2}& \mathrm{2b\; +\; 14}\end{array},,\right)\right)$
   
   2a = 4
   a = 2
   
   2b = 4a + 2
   2b = 4(2) + 2
   2b = 8 + 2
   2b = 10
   b = 5
   
   3c = 2b + 14
   3c = 2(5) + 14
   3c = 10 + 14
   3c = 24
   c = 8
   


3. 1. AB = $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{3\; .\; -1\; +\; 2\; .\; 5}& \mathrm{3\; .\; 4\; +\; 2\; .\; 0}\\ \mathrm{1\; .\; -1\; +\; -1\; .\; 5}& \mathrm{1\; .\; 4\; +\; -1\; .\; 0}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}7& 12\\ -6& 4\end{array},,\right)\right)$
   
   
   2. AC = $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{3\; .\; 1\; +\; 2\; .\; -2}& \mathrm{3\; .\; -3\; +\; 2\; .\; 5}& \mathrm{3\; .\; 0\; +\; 2\; .\; 1}\\ \mathrm{1\; .\; 1\; +\; -1\; .\; -2}& \mathrm{1\; .\; -3\; +\; -1\; .\; 5}& \mathrm{1\; .\; 0\; +\; -1\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-1& 1& 2\\ 3& -8& -1\end{array},,\right)\right)$
   
   
   3. BC = $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{-1\; .\; 1\; +\; 4\; .\; -2}& \mathrm{-1\; .\; -3\; +\; 4\; .\; 5}& \mathrm{-1\; .\; 0\; +\; 4\; .\; 1}\\ \mathrm{5\; .\; 1\; +\; 0\; .\; -2}& \mathrm{5\; .\; -3\; +\; 0\; .\; 5}& \mathrm{5\; .\; 0\; +\; 0\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-9& 23& 4\\ 5& -15& 0\end{array},,\right)\right)$
   
   
   4. A^t = $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$
      
      A^tC = $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{3\; .\; 1\; +\; 1\; .\; -2}& \mathrm{3\; .\; -3\; +\; 1\; .\; 5}& \mathrm{3\; .\; 0\; +\; 1\; .\; 1}\\ \mathrm{2\; .\; 1\; +\; -1\; .\; -2}& \mathrm{2\; .\; -3\; +\; -1\; .\; 5}& \mathrm{2\; .\; 0\; +\; -1\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}1& -4& 1\\ 4& -11& -1\end{array},,\right)\right)$
   
   
   5. B^t = $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$
      
      B^tC = $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{-1\; .\; 1\; +\; 5\; .\; -2}& \mathrm{-1\; .\; -3\; +\; 5\; .\; 5}& \mathrm{-1\; .\; 0\; +\; 5\; .\; 1}\\ \mathrm{4\; .\; 1\; +\; 0\; .\; -2}& \mathrm{4\; .\; -3\; +\; 0\; .\; 5}& \mathrm{4\; .\; 0\; +\; 0\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-11& 28& 5\\ 4& -12& 0\end{array},,\right)\right)$
   
   
   6. C^t = $\left(\left(,,\begin{array}{cc}1& -2\\ -3& 5\\ 0& 1\end{array},,\right)\right)$
      
      C^tA = $\left(\left(,,\begin{array}{cc}1& -2\\ -3& 5\\ 0& 1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{1\; .\; 3\; +\; -2\; .\; 1}& \mathrm{1\; .\; 2\; +\; -2\; .\; -1}\\ \mathrm{-3\; .\; 3\; +\; 5\; .\; 1}& \mathrm{-3\; .\; 2\; +\; 5\; .\; -1}\\ \mathrm{0\; .\; 3\; +\; 1\; .\; 1}& \mathrm{0\; .\; 2\; +\; 1\; .\; -1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}1& 4\\ -4& -11\\ 1& -1\end{array},,\right)\right)$
   
   
   7. C^tB = $\left(\left(,,\begin{array}{cc}1& -2\\ -3& 5\\ 0& 1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{1\; .\; -1\; +\; -2\; .\; 5}& \mathrm{1\; .\; 4\; +\; -2\; .\; 0}\\ \mathrm{-3\; .\; -1\; +\; 5\; .\; 5}& \mathrm{-3\; .\; 4\; +\; 5\; .\; 0}\\ \mathrm{0\; .\; -1\; +\; 1\; .\; 5}& \mathrm{0\; .\; 4\; +\; 1\; .\; 0}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}-11& 4\\ 28& -12\\ 5& 0\end{array},,\right)\right)$
   
   
   8. A^tB^t = $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{3\; .\; -1\; +\; 1\; .\; 4}& \mathrm{3\; .\; 5\; +\; 1\; .\; 0}\\ \mathrm{2\; .\; -1\; +\; -1\; .\; 4}& \mathrm{2\; .\; 5\; +\; -1\; .\; 0}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}1& 15\\ -6& 10\end{array},,\right)\right)$
   
   
   9. B^tA^t = $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{-1\; .\; 3\; +\; 5\; .\; 2}& \mathrm{-1\; .\; 1\; +\; 5\; .\; -1}\\ \mathrm{4\; .\; 3\; +\; 0\; .\; 2}& \mathrm{4\; .\; 1\; +\; 0\; .\; -1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}7& -6\\ 12& 4\end{array},,\right)\right)$
   
   
   10. AB = $\left(\left(,,\begin{array}{cc}7& 12\\ -6& 4\end{array},,\right)\right)$
       
       (AB)^t = $\left(\left(,,\begin{array}{cc}7& -6\\ 12& 4\end{array},,\right)\right)$
   
   
   11. Apakah (AB)^t = B^tA^t ?
       Iya


4. $\left(\left(,,\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 3& -2\end{array},,\right)\right)\left(\left(,,\begin{array}{cc}6& -5\\ 2& 4\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a\; .\; 6\; +\; b\; .\; 2}& \mathrm{a\; .\; -5\; +\; b\; .\; 4}\\ \mathrm{3\; .\; 6\; +\; -2\; .\; 2}& \mathrm{3\; .\; -5\; +\; -2\; .\; 4}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{cc}\mathrm{6a\; +\; 2b}& \mathrm{-5a\; +\; 4b}\\ 14& -23\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   
   6a + 2b = 12
   2b = 12 - 6a
   b = 6 - 3a
   
   -5a + 4b = -27
   -5a + 4(6 - 3a) = -27
   -5a + 24 - 12a = -27
   -17a = -27 - 24
   -17a = -51
   a = 3
   
   b = 6 - 3a
   b = 6 - 3(3)
   b = 6 - 9
   b = -3


5. $\left(\left(,,\begin{array}{cc}1& 5\\ 4& -6\end{array},,\right)\right)\left(\left(,,\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{c}\mathrm{1\; .\; x\; +\; 5\; .\; y}\\ \mathrm{4\; .\; x\; +\; -6\; .\; y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{c}\mathrm{x\; +\; 5y}\\ \mathrm{4x\; -\; 6y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   x + 5y = -13
   x = -13 - 5y
   
   4x - 6y = 26
   4(-13 - 5y) - 6y = 26
   -52 - 20y - 6y = 26
   -26y = 26 + 52
   -26y = 78
   y = -3
   
   x = -13 - 5y
   x = -13 - 5(-3)
   x = -13 + 15
   x = 2

Jawaban Soal Latihan 1 Penjumlahan Dan Pengurangan Matriks

Soal Latihan 1

1. Diketahui matriks A, B dan C sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)$
   
   C = $\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)$
   
   Tentukan :
   1. A + B
   2. A - B
   3. B + C
   4. A - C
   5. C - B


2. Tentukan nilai x, y dan z dari persamaan matriks berikut :
   
   $\left(\left(,,\begin{array}{cc}\mathrm{x}& \mathrm{2y}\\ 8& \mathrm{3z}\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}5& 3\\ 1& 6\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}10& 11\\ 9& 15\end{array},,\right)\right)$
   
   


3. Diketahui matriks A dan B sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}6& \mathrm{z}\\ 10& \mathrm{x}\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}\mathrm{2x}& \mathrm{2x}\\ 10& \frac{3}{2}\mathrm{y}\end{array},,\right)\right)$
   
   Tentukan nilai x, y dan z jika A - B = O

Jawab :

1. 1. A + B = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}1& 3\\ 3& -2\end{array},,\right)\right)$
      
   
   
   2. A - B = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-7& 1\\ -1& 8\end{array},,\right)\right)$
      
      
   3. B + C = $\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}6& -1\\ 5& -6\end{array},,\right)\right)$
      
      
   4. A - C = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-5& 4\\ -2& 4\end{array},,\right)\right)$
      
      
   5. C - B = $\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-2& -3\\ 1& 4\end{array},,\right)\right)$
      
      


2. $\left(\left(,,\begin{array}{cc}\mathrm{x}& \mathrm{2y}\\ 8& \mathrm{3z}\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}5& 3\\ 1& 6\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}10& 11\\ 9& 15\end{array},,\right)\right)$
   
   x + 5 = 10
   x = 10 - 5
   x = 5
   
   2y + 3 = 11
   2y = 11 - 3
   2y = 8
   y = 4
   
   3z + 6 = 15
   3z = 15 - 6
   3z = 9
   z = 3


3. A - B = O
   
   $\left(\left(,,\begin{array}{cc}6& \mathrm{z}\\ 10& \mathrm{x}\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}\mathrm{2x}& \mathrm{2x}\\ 10& \frac{3}{2}\mathrm{y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}0& 0\\ 0& 0\end{array},,\right)\right)$
   
   6 - 2x = 0
   6 = 2x
   x = 3
   
   z - 2x = 0
   z - 2(3) = 0
   z = 6
   
   $\mathrm{x}-\frac{3}{2}\mathrm{y}=0$
   $3-\frac{3}{2}\mathrm{y}=0$
   $3=\frac{3}{2}\mathrm{y}$
   $\mathrm{y}=3.\frac{2}{3}$
   $\mathrm{y}=2$
   

Jawaban Soal Latihan 1 Matriks

Soal Latihan 1

1. Diketahui matriks A sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{ccccc}9& 0& 3& -7& 6\\ -5& 6& 8& 1& 4\\ 5& 7& 0& 2& 8\end{array},,\right)\right)$
   
   
   1. Sebutkan ordo dari matriks A
   2. Sebutkan elemen-elemen kolom ke 2
   3. Sebutkan elemen-elemen baris ke 3
   4. Jika a_ij menyatakan elemen baris ke i dan kolom ke j maka tentukan a₂₃, a₁₄ dan a₃₅


2. Berikan contoh matriks berordo 3x3 untuk :
   1. Matriks diagonal
   2. Matriks identitas
   3. Matriks segitiga atas
   4. Matriks segitiga bawah


3. Tulislah transpose dari matriks berikut :
   
   
   1. A = $\left(\left(,,\begin{array}{cc}4& 3\\ 5& 6\end{array},,\right)\right)$
   
   
   2. B = $\left(\left(,,\begin{array}{cc}2& 4\\ -3& 1\\ 0& 5\end{array},,\right)\right)$
   
   
   3. C = $\left(\left(,,\begin{array}{ccc}4& 5& -1\\ 2& 3& 0\end{array},,\right)\right)$
   
   
   4. D = $\left(\left(,,\begin{array}{ccc}4& 0& 5\\ 2& 4& 7\\ -1& 3& -2\end{array},,\right)\right)$


4. Tentukan nilai x dan y dari persamaan berikut :
   
   
   1. $\left(\left(,,\begin{array}{cc}\mathrm{x}& \mathrm{2y}\\ 0& 3\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}1& 8\\ 0& 3\end{array},,\right)\right)$
   
   
   2. $\left(\left(,,\begin{array}{cc}\mathrm{x\; +\; y}& 5\\ 4& \mathrm{x\; -\; y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}5& 5\\ 4& 1\end{array},,\right)\right)$
   
   
   3. $\left(\left(,,\begin{array}{cc}\mathrm{x}& 4\\ \mathrm{2y}& \mathrm{3z}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}\mathrm{4z\; -\; 6y}& \mathrm{2x}\\ \mathrm{4x\; +\; 2}& \mathrm{2y\; +\; 14}\end{array},,\right)\right)$


5. Tentukan x dan y jika A = B^t
   
   
   1. A = $\left(\left(,,\begin{array}{cc}\mathrm{2x}& 0\\ 1& \mathrm{-4y}\end{array},,\right)\right)$ dan B = $\left(\left(,,\begin{array}{cc}8& 1\\ 0& 40\end{array},,\right)\right)$
   
   
   2. A = $\left(\left(,,\begin{array}{cc}\mathrm{x\; -\; 2y}& 1\\ 0& \mathrm{-3y}\end{array},,\right)\right)$ dan B = $\left(\left(,,\begin{array}{cc}8& 0\\ 1& 40\end{array},,\right)\right)$

Jawab :

1. 1. ordo mariks A = 3 x 5
   2. elemen-elemen kolom ke 2 = 0, 6, 7
   3. elemen-elemen baris ke 3 = 5, 7, 0, 2, 8
   4. a₂₃ = 8
      a₁₄ = -7
      a₃₅ = 8


2. 1. matriks diagonal = $\left(\left(,,\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array},,\right)\right)$
   
   
   2. matriks identitas = $\left(\left(,,\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array},,\right)\right)$
   
   
   3. matriks segitiga atas = $\left(\left(,,\begin{array}{ccc}4& 1& 5\\ 0& 2& 7\\ 0& 0& 3\end{array},,\right)\right)$
   
   
   4. matriks segitiga bawah = $\left(\left(,,\begin{array}{ccc}4& 0& 0\\ 2& 6& 0\\ 5& 3& 1\end{array},,\right)\right)$


3. 1. A^t = $\left(\left(,,\begin{array}{cc}4& 5\\ 3& 6\end{array},,\right)\right)$
   
   
   2. B^t = $\left(\left(,,\begin{array}{ccc}2& -3& 0\\ 4& 1& 5\end{array},,\right)\right)$
   
   
   3. C^t = $\left(\left(,,\begin{array}{cc}4& 2\\ 5& 3\\ -1& 0\end{array},,\right)\right)$
   
   
   4. D^t = $\left(\left(,,\begin{array}{ccc}4& 2& -1\\ 0& 4& 3\\ 5& 7& -2\end{array},,\right)\right)$


4. 1. x = 1    2y = 8
             y = 4
   
   
   2. x + y = 5
      x = 5 - y
      
      x - y = 1
      (5 - y) - y = 1
      5 - 2y = 1
      -2y = 1 - 5
      -2y = -4
      y = 2
      
      x = 5 - y
      x = 5 - 2
      x = 3
   
   
   3. 4 = 2x
      x = 2
      
      2y = 4x + 2
      2y = 4(2) + 2
      2y = 8 + 2
      2y = 10
      y = 5
      
      3z = 2y + 14
      3z = 2(5) + 14
      3z = 10 + 14
      3z = 24
      z = 8


5. 1. B^t = $\left(\left(,,\begin{array}{cc}8& 0\\ 1& 40\end{array},,\right)\right)$
      
      A = B^t
      $\left(\left(,,\begin{array}{cc}\mathrm{2x}& 0\\ 1& \mathrm{-4y}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}8& 0\\ 1& 40\end{array},,\right)\right)$
      
      2x = 8
      x = 4
      
      -4y = 40
      y = -10
   
   
   2. B^t = $\left(\left(,,\begin{array}{cc}8& 1\\ 0& 40\end{array},,\right)\right)$
      
      A = B^t
      $\left(\left(,,\begin{array}{cc}\mathrm{x\; -\; 2y}& 1\\ 0& \mathrm{-3y}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}8& 1\\ 0& 40\end{array},,\right)\right)$
      
      -3y = 40
      $\mathrm{y}=-\frac{40}{3}$
      
      x - 2y = 8
      $\mathrm{x}-2(-\frac{40}{3})=8$
      $\mathrm{x}+\frac{80}{3}=8$
      $\mathrm{x}=8-\frac{80}{3}$
      $\mathrm{x}=\frac{24}{3}-\frac{80}{3}$
      $\mathrm{x}=-\frac{56}{3}$
      

Jawaban Soal Latihan 1 Perbandingan Trigonometri Suatu Sudut Dalam Segitiga Siku-siku

Soal Latihan 1

1. Diketahui segitiga ABC sebagai berikut :
   
   
   
   
   Tentukan :
   
   
   1. Panjang AC
   2. Nilai perbandingan : sin α, cos α, tan α, cot α, sec α, cosec α






2. Pada segitiga KLM siku-siku di K, diketahui nilai cos M = $\frac{1}{5}\sqrt{5}$. Tentukan nilai tan M dan cosec M
3. Ditentukan bahwa a° adalah besar sudut XOP. Hitunglah nilai dari sin a°, cos a°, tan a°, cot a°, sec a° dan cosec a°, jika koordinat titik P adalah :
   1. (8, 6)
   2. (-4, 6)
   3. (12, -5)
   4. (-3, -4)
   5. (1, 3)

Jawab :

1. 1. AB = 12
      BC = 5
      $AC=\sqrt{{AB}^2+{BC}^2}$
      $=\sqrt{{12}^2+5^2}$
      $=\sqrt{144+25}$
      $=\sqrt{169}$
      $=13$
      
   2. sin α = $\frac{sisi\; hadap}{sisi\; miring}=\frac{5}{13}$
      cos α = $\frac{sisi\; dekat}{sisi\; miring}=\frac{12}{13}$
      tan α = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{5}{12}$
      cot α = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{12}{5}$
      sec α = $\frac{sisi\; miring}{sisi\; dekat}=\frac{13}{12}$
      cosec α = $\frac{sisi\; miring}{sisi\; hadap}=\frac{13}{5}$
2. cos M = $\frac{1}{5}\sqrt{5}$
   
   
   
   
   tan M = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{2\sqrt{5}}{\sqrt{5}}=2$
   
   cosec M = $\frac{sisi\; miring}{sisi\; hadap}=\frac{5}{2\sqrt{5}}=\frac{5}{2\sqrt{5}}.\frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{10}=\frac{1}{2}\sqrt{5}$







3. 1. 
      
      sisi hadap = 6
      sisi dekat = 8
      sisi miring = $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{6}{10}=\frac{3}{5}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{8}{10}=\frac{4}{5}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{6}{8}=\frac{3}{4}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{8}{6}=\frac{4}{3}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{10}{8}=\frac{5}{4}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{10}{6}=\frac{5}{3}$
      
   
   
   
   2. 
      
      sisi hadap = 6
      sisi dekat = -4
      sisi miring = $\sqrt{6^2+{\mathrm{(-4)}}^2}=\sqrt{36+16}=\sqrt{52}=\sqrt{4.13}=2\sqrt{13}$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{6}{2\sqrt{13}}=\frac{3}{\sqrt{13}}.\frac{\sqrt{13}}{\sqrt{13}}=\frac{3\sqrt{13}}{13}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{-4}{2\sqrt{13}}=\frac{-2}{\sqrt{13}}.\frac{\sqrt{13}}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{6}{-4}=-\frac{3}{2}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{-4}{6}=-\frac{2}{3}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{2\sqrt{13}}{-4}=-\frac{1}{2}\sqrt{13}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{2\sqrt{13}}{6}=\frac{1}{3}\sqrt{13}$
      
   
   
   
   
   3. 
      sisi hadap = -5
      sisi dekat = 12
      sisi miring = $\sqrt{{\mathrm{(-5)}}^2+{12}^2}=\sqrt{25+144}=\sqrt{169}=13$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{-5}{13}=-\frac{5}{13}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{12}{13}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{-5}{12}=-\frac{5}{12}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{12}{-5}=-\frac{12}{5}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{13}{12}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{13}{-5}=-\frac{13}{5}$
      
   
   
   4. 
      
      
      sisi hadap = -4
      sisi dekat = -3
      sisi miring = $\sqrt{{\mathrm{(-4)}}^2+{\mathrm{(-3)}}^2}=\sqrt{16+9}=\sqrt{25}=5$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{-4}{5}=-\frac{4}{5}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{-3}{5}=-\frac{3}{5}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{-4}{-3}=\frac{4}{3}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{-3}{-4}=\frac{3}{4}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{5}{-3}=-\frac{5}{3}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{5}{-4}=-\frac{5}{4}$
      
   
   
   
   
   
   5. 
      
      
      sisi hadap = 3
      sisi dekat = 1
      sisi miring = $\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{3}{\sqrt{10}}=\frac{3}{\sqrt{10}}.\frac{\sqrt{10}}{\sqrt{10}}=\frac{3\sqrt{10}}{10}=\frac{3}{10}\sqrt{10}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}.\frac{\sqrt{10}}{\sqrt{10}}=\frac{1}{10}\sqrt{10}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{3}{1}=3$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{1}{3}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{\sqrt{10}}{1}=\sqrt{10}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{\sqrt{10}}{3}=\frac{1}{3}\sqrt{10}$

Jawaban Soal Latihan 1 Rumus Trigonometri Penjumlahan Dan Pengurangan Sinus Dan Cosinus

Soal Latihan 1

1. Sederhanakan :
   1. $\frac{Cos\; 75°\; -\; Cos\; 15°}{Sin\; 75°\; +\; Sin\; 15°}$
   2. $\frac{Sin\; 75°\; -\; Cos\; 75°}{Cos\; 75°\; +\; Cos\; 15°}$
   3. $\frac{Cos\; 105°\; +\; Cos\; 15°}{Sin\; 105°\; -\; Sin\; 15°}$
2. Buktikan identitas berikut :
   1. $\frac{Sin\; 7a\; -\; Sin\; 5a}{Cos\; 7a\; +\; Cos\; 5a}=\mathrm{Tan\; a}$
   2. $\frac{Sin\; a\; +\; Sin\; 3a}{Cos\; a\; -\; Cos\; 3a}=\mathrm{Cot\; a}$
   3. $\frac{Sin\; a\; +\; Sin\; 2a}{Cos\; a\; -\; Cos\; 2a}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
3. Buktikan :
   1. $\frac{Sin\; 4A\; +\; Sin\; 2A}{Cos\; 4A\; +\; Cos\; 2A}=\mathrm{Tan\; 3A}$
   2. $\frac{Cos\; 3A\; -\; Cos\; 5A}{Sin\; 3A\; -\; Sin\; A}=\mathrm{2\; Sin\; 2A}$
4. Jika k = Sin a + Sin b dan m = Cos a + Cos b. Buktikan :
   1. $\mathrm{k}+\mathrm{m}=2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}$
   2. $\mathrm{k}=\mathrm{m}\mathrm{Tan}\frac{1}{2}(\mathrm{a}+\mathrm{b})$

Jawab :

1. 1. $\frac{Cos\; 75°\; -\; Cos\; 15°}{Sin\; 75°\; +\; Sin\; 15°}$
      $=\frac{-2\; Sin\left(\frac{75°\; +\; 15°}{2}\right).\; Sin\left(\frac{75°\; -\; 15°}{2}\right)}{2\; Sin\left(\frac{75°\; +\; 15°}{2}\right).\; Cos\left(\frac{75°\; +\; 15°}{2}\right)}$
      $=\frac{-2\; Sin\; 45°\; .\; Sin\; 30°}{2\; Sin\; 45°\; .\; Cos\; 30°}$
      $=\frac{-2.\frac{1}{2}\sqrt{2}.\frac{1}{2}}{2.\frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}}$
      $=\frac{-\frac{1}{2}\sqrt{2}}{\frac{1}{2}\sqrt{2}\sqrt{3}}$
      $=-\frac{1}{\sqrt{3}}$
      $=-\frac{1}{3}\sqrt{3}$
      
   
   
   2. $\frac{Sin\; 75°\; -\; Cos\; 75°}{Cos\; 75°\; +\; Cos\; 15°}$
      (Dengan Menggunakan Rumus Trigonometri Sudut Berelasi)
      $=\frac{Sin\; 75°\; -\; Cos\; (90°\; -\; 15°)}{Cos\; 75°\; +\; Cos\; 15°}$
      $=\frac{Sin\; 75°\; -\; Sin\; 15°}{Cos\; 75°\; +\; Cos\; 15°}$
      $=\frac{2\; Cos\left(\frac{75°\; +\; 15°}{2}\right).\; Sin\left(\frac{75°\; -\; 15°}{2}\right)}{2\; Cos\left(\frac{75°\; +\; 15°}{2}\right).\; Cos\left(\frac{75°\; -\; 15°}{2}\right)}$
      $=\frac{2\; Cos\; 45°\; .\; Sin\; 30°}{2\; Cos\; 45°\; .\; Cos\; 30°}$
      $=\frac{2.\frac{1}{2}\sqrt{2}.\frac{1}{2}}{2.\frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}}$
      $=\frac{1}{\sqrt{3}}$
      $=\frac{1}{3}\sqrt{3}$
      
   
   
   3. $\frac{Cos\; 105°\; +\; Cos\; 15°}{Sin\; 105°\; -\; Sin\; 15°}$
      $=\frac{2\; Cos\left(\frac{105°\; +\; 15°}{2}\right).\; Cos\left(\frac{105°\; -\; 15°}{2}\right)}{2\; Cos\left(\frac{105°\; +\; 15°}{2}\right).\; Sin\left(\frac{105°\; -\; 15°}{2}\right)}$
      $=\frac{2\; Cos\; 60°\; .\; Cos\; 45°}{2\; Cos\; 60°\; .\; Sin\; 45°}$
      $=\frac{2\; .\frac{1}{2}.\frac{1}{2}\sqrt{2}}{2\; .\frac{1}{2}.\frac{1}{2}\sqrt{2}}$
      $=1$
      


2. 1. $\frac{Sin\; 7a\; -\; Sin\; 5a}{Cos\; 7a\; +\; Cos\; 5a}=\mathrm{Tan\; a}$
      $\frac{2\; Cos\left(\frac{7a\; +\; 5a}{2}\right).\; Sin\left(\frac{7a\; -\; 5a}{2}\right)}{2\; Cos\left(\frac{7a\; +\; 5a}{2}\right).\; Cos\left(\frac{7a\; -\; 5a}{2}\right)}=\mathrm{Tan\; a}$
      $\frac{2\; Cos\; 6a\; .\; Sin\; a}{2\; Cos\; 6a\; .\; Cos\; a}=\mathrm{Tan\; a}$
      $\frac{Sin\; a}{Cos\; a}=\mathrm{Tan\; a}$
      (Dengan Menggunakan Identitas Trigonometri)
      $\mathrm{Tan\; a}=\mathrm{Tan\; a}$
      
   
   
   2. $\frac{Sin\; a\; +\; Sin\; 3a}{Cos\; a\; -\; Cos\; 3a}=\mathrm{Cot\; a}$
      $\frac{2\; Sin\left(\frac{a\; +\; 3a}{2}\right).\; Cos\left(\frac{a\; -\; 3a}{2}\right)}{-2\; Sin\left(\frac{a\; +\; 3a}{2}\right).\; Sin\left(\frac{a\; -\; 3a}{2}\right)}=\mathrm{Cot\; a}$
      $\frac{2\; Sin\; 2a\; .\; Cos\; (-a)}{-2\; Sin\; 2a\; .\; Sin\; (-a)}=\mathrm{Cot\; a}$
      (Dengan Menggunakan Rumus Trigonometri Sudut Berelasi)
      $\frac{Cos\; a}{-(-Sin\; a)}=\mathrm{Cot\; a}$
      $\frac{Cos\; a}{Sin\; a}=\mathrm{Cot\; a}$
      (Dengan Menggunakan Identitas Trigonometri)
      $\mathrm{Cot\; a}=\mathrm{Cot\; a}$
      
   
   
   3. $\frac{Sin\; a\; +\; Sin\; 2a}{Cos\; a\; -\; Cos\; 2a}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
      $\frac{2\; Sin\left(\frac{a\; +\; 2a}{2}\right).\; Cos\left(\frac{a\; -\; 2a}{2}\right)}{-2\; Sin\left(\frac{a\; +\; 2a}{2}\right).\; Sin\left(\frac{a\; -\; 2a}{2}\right)}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
      $\frac{2\; Sin\frac{3}{2}a\; .\; Cos(-\frac{1}{2}\mathrm{a})}{-2\; Sin\frac{3}{2}a\; .\; Sin(-\frac{1}{2}\mathrm{a})}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
      (Dengan Menggunakan Rumus Trigonometri Sudut Berelasi)
      $\frac{Cos\frac{1}{2}\mathrm{a}}{-(-Sin\frac{1}{2}\mathrm{a})}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
      $\frac{Cos\frac{1}{2}\mathrm{a}}{Sin\frac{1}{2}\mathrm{a}}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
      (Dengan Menggunakan Identitas Trigonometri)
      $\mathrm{Cot}\frac{1}{2}\mathrm{a}=\mathrm{Cot}\frac{1}{2}\mathrm{a}$
      


3. 1. $\frac{Sin\; 4A\; +\; Sin\; 2A}{Cos\; 4A\; +\; Cos\; 2A}=\mathrm{Tan\; 3A}$
      $\frac{2\; Sin\left(\frac{4A\; +\; 2A}{2}\right).\; Cos\left(\frac{4A\; -\; 2A}{2}\right)}{2\; Cos\left(\frac{4A\; +\; 2A}{2}\right).\; Cos\left(\frac{4A\; -\; 2A}{2}\right)}=\mathrm{Tan\; 3A}$
      $\frac{2\; Sin\; 3A\; .\; Cos\; A}{2\; Cos\; 3A\; .\; Cos\; A}=\mathrm{Tan\; 3A}$
      $\frac{Sin\; 3A}{Cos\; 3A}=\mathrm{Tan\; 3A}$
      (Dengan Menggunakan Identitas Trigonometri)
      $\mathrm{Tan\; 3A}=\mathrm{Tan\; 3A}$
      
   
   
   2. $\frac{Cos\; 3A\; -\; Cos\; 5A}{Sin\; 3A\; -\; Sin\; A}=\mathrm{2\; Sin\; 2A}$
      $\frac{-2\; Sin\left(\frac{3A\; +\; 5A}{2}\right).\; Sin\left(\frac{3A\; -\; 5A}{2}\right)}{2\; Cos\left(\frac{3A\; +\; A}{2}\right).\; Sin\left(\frac{3A\; -\; A}{2}\right)}=\mathrm{2\; Sin\; 2A}$
      $\frac{-2\; Sin\; 4A\; .\; Sin\; (-A)}{2\; Cos\; 2A\; .\; Sin\; A}=\mathrm{2\; Sin\; 2A}$
      (Dengan Menggunakan Rumus Trigonometri Sudut Berelasi)
      $\frac{-2\; Sin\; 4A\; .\; (-Sin\; A)}{2\; Cos\; 2A\; .\; Sin\; A}=\mathrm{2\; Sin\; 2A}$
      $\frac{2\; Sin\; 4A\; .\; Sin\; A}{2\; Cos\; 2A\; .\; Sin\; A}=\mathrm{2\; Sin\; 2A}$
      $\frac{Sin\; 4A}{Cos\; 2A}=\mathrm{2\; Sin\; 2A}$
      $\frac{Sin\; 2(2A)}{Cos\; 2A}=\mathrm{2\; Sin\; 2A}$
      (Dengan Menggunakan Rumus Trigonometri Sudut Rangkap)
      $\frac{2\; Sin\; 2A\; .\; Cos\; 2A}{Cos\; 2A}=\mathrm{2\; Sin\; 2A}$
      $\mathrm{2\; Sin\; 2A}=\mathrm{2\; Sin\; 2A}$
      


4. k = Sin a + Sin b
   m = Cos a + Cos b
   
   
   1. $\mathrm{k}+\mathrm{m}=2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}$
      $\mathrm{(Sin\; a\; +\; Sin\; b)}+\mathrm{(Cos\; a\; +\; Cos\; b)}=2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}$
      $2.\mathrm{Sin}\left(\frac{a\; +\; b}{2}\right).\mathrm{Cos}\left(\frac{a\; -\; b}{2}\right)+2.\mathrm{Cos}\left(\frac{a\; +\; b}{2}\right).\mathrm{Cos}\left(\frac{a\; -\; b}{2}\right)=2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}$
      $2.\mathrm{Sin}\frac{1}{2}(\mathrm{a}+\mathrm{b}).\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})+2.\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}).\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})=2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}$
      $2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}=2\mathrm{Cos}\frac{1}{2}(\mathrm{a}-\mathrm{b})\left\{\mathrm{Sin}\frac{1}{2}\right(\mathrm{a}+\mathrm{b})+\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b}\left)\right\}$
      
   
   
   2. $\mathrm{k}=\mathrm{m}\mathrm{Tan}\frac{1}{2}(\mathrm{a}+\mathrm{b})$
      $\mathrm{Sin\; a\; +\; Sin\; b}=\mathrm{(Cos\; a\; +\; Cos\; b)}.\mathrm{Tan}\frac{1}{2}(\mathrm{a}+\mathrm{b})$
      $2.\mathrm{Sin}\left(\frac{a\; +\; b}{2}\right).\mathrm{Cos}\left(\frac{a\; -\; b}{2}\right)=\{2.\mathrm{Cos}(\frac{a\; +\; b}{2}).\mathrm{Cos}(\frac{a\; -\; b}{2}\left)\right\}.\mathrm{Tan}\frac{1}{2}(\mathrm{a}+\mathrm{b})$
      $2.\mathrm{Sin}\frac{1}{2}\mathrm{(a\; +\; b)}.\mathrm{Cos}\frac{1}{2}\mathrm{(a\; -\; b)}=2.\mathrm{Cos}\frac{1}{2}\mathrm{(a\; +\; b)}.\mathrm{Cos}\frac{1}{2}\mathrm{(a\; -\; b)}.\frac{\mathrm{Sin}\frac{1}{2}(\mathrm{a}+\mathrm{b})}{\mathrm{Cos}\frac{1}{2}(\mathrm{a}+\mathrm{b})}$
      $2.\mathrm{Sin}\frac{1}{2}\mathrm{(a\; +\; b)}.\mathrm{Cos}\frac{1}{2}\mathrm{(a\; -\; b)}=2.\mathrm{Sin}\frac{1}{2}\mathrm{(a\; +\; b)}.\mathrm{Cos}\frac{1}{2}\mathrm{(a\; -\; b)}.$