If sin A = 12/13 and sin B = 4/5, Find sin(A+B) and cos(A+B)
Question
If \[ \sin A=\frac{12}{13} \] and \[ \sin B=\frac{4}{5} \] where \[ \frac{\pi}{2} < A < \pi \] and \[ 0 < B < \frac{\pi}{2} \] find:
(i) \(\sin(A+B)\)
(ii) \(\cos(A+B)\)
Solution
Given: \[ \sin A=\frac{12}{13} \]
Using \[ \sin^2 A+\cos^2 A=1 \]
\[ \cos A=\sqrt{1-\left(\frac{12}{13}\right)^2} \]
\[ =\sqrt{1-\frac{144}{169}} \]
\[ =\sqrt{\frac{25}{169}} \]
\[ \cos A=\frac{5}{13} \]
Since \[ \frac{\pi}{2} < A < \pi \] A lies in the second quadrant, where cosine is negative.
Therefore, \[ \cos A=-\frac{5}{13} \]
Also, \[ \sin B=\frac{4}{5} \]
Using \[ \sin^2 B+\cos^2 B=1 \]
\[ \cos B=\sqrt{1-\left(\frac{4}{5}\right)^2} \]
\[ =\sqrt{1-\frac{16}{25}} \]
\[ =\sqrt{\frac{9}{25}} \]
\[ \cos B=\frac{3}{5} \]
Since \[ 0 < B < \frac{\pi}{2} \] B lies in the first quadrant, so cosine is positive.
(i) Find \(\sin(A+B)\)
Using formula: \[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]
\[ =\frac{12}{13}\times\frac{3}{5}+\left(-\frac{5}{13}\right)\times\frac{4}{5} \]
\[ =\frac{36}{65}-\frac{20}{65} \]
\[ =\frac{16}{65} \]
Therefore, \[ \boxed{\sin(A+B)=\frac{16}{65}} \]
(ii) Find \(\cos(A+B)\)
Using formula: \[ \cos(A+B)=\cos A\cos B-\sin A\sin B \]
\[ =\left(-\frac{5}{13}\right)\times\frac{3}{5}-\frac{12}{13}\times\frac{4}{5} \]
\[ =-\frac{15}{65}-\frac{48}{65} \]
\[ =-\frac{63}{65} \]
Therefore, \[ \boxed{\cos(A+B)=-\frac{63}{65}} \]