If cos A = -24/25 and cos B = 3/5, Find sin(A+B) and cos(A+B)

Question

If \[ \cos A=-\frac{24}{25} \] and \[ \cos B=\frac{3}{5} \] where \[ \pi < A < \frac{3\pi}{2} \] and \[ \frac{3\pi}{2} < B < 2\pi \] find:

(i) \(\sin(A+B)\)
(ii) \(\cos(A+B)\)

Solution

Given: \[ \cos A=-\frac{24}{25} \]

Using \[ \sin^2 A+\cos^2 A=1 \]

\[ \sin A=\sqrt{1-\left(-\frac{24}{25}\right)^2} \]

\[ =\sqrt{1-\frac{576}{625}} \]

\[ =\sqrt{\frac{49}{625}} \]

\[ \sin A=\frac{7}{25} \]

Since \[ \pi < A < \frac{3\pi}{2} \] A lies in the third quadrant, so sine is negative.

Therefore, \[ \sin A=-\frac{7}{25} \]

Also, \[ \cos B=\frac{3}{5} \]

Using \[ \sin^2 B+\cos^2 B=1 \]

\[ \sin B=\sqrt{1-\left(\frac{3}{5}\right)^2} \]

\[ =\sqrt{1-\frac{9}{25}} \]

\[ =\sqrt{\frac{16}{25}} \]

\[ \sin B=\frac{4}{5} \]

Since \[ \frac{3\pi}{2} < B < 2\pi \] B lies in the fourth quadrant, so sine is negative.

Therefore, \[ \sin B=-\frac{4}{5} \]

(i) Find \(\sin(A+B)\)

Using formula: \[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]

\[ =\left(-\frac{7}{25}\right)\times\frac{3}{5}+\left(-\frac{24}{25}\right)\times\left(-\frac{4}{5}\right) \]

\[ =-\frac{21}{125}+\frac{96}{125} \]

\[ =\frac{75}{125} \]

\[ =\frac{3}{5} \]

Therefore, \[ \boxed{\sin(A+B)=\frac{3}{5}} \]

(ii) Find \(\cos(A+B)\)

Using formula: \[ \cos(A+B)=\cos A\cos B-\sin A\sin B \]

\[ =\left(-\frac{24}{25}\right)\times\frac{3}{5}-\left(-\frac{7}{25}\right)\times\left(-\frac{4}{5}\right) \]

\[ =-\frac{72}{125}-\frac{28}{125} \]

\[ =-\frac{100}{125} \]

\[ =-\frac{4}{5} \]

Therefore, \[ \boxed{\cos(A+B)=-\frac{4}{5}} \]