Question

If $sin\left(A\right)=\frac{-8}{17},\frac{3\pi }{2}<A<2\pi ,$and $cos\left(B\right)=\frac{-24}{25},\pi <B<\frac{3\pi }{2},cos(2A+B)=$

• A. $\frac{-5544}{7225}$
• B. $\frac{-2184}{7225}$
• C. $\frac{-3696}{7225}$
• D. $\frac{2184}{7225}$
• E. $\frac{5544}{7225}$

Answer 1

The correct option is A $\frac{-5544}{7225}$

$\sin \left(A\right)=-\frac{8}{17}$, $\frac{3\pi }{2}<A<2\pi$ gives $\cos \left(A\right)=\frac{15}{17}$.

While $\cos \left(B\right)=\frac{-24}{25}$, $\pi <B<\frac{3\pi }{2}$ gives $\sin \left(B\right)=-\frac{-7}{25}$.

Since, $\cos \left(2A\right)={\cos }^2\left(A\right)-{\sin }^2\left(A\right)$ and $\sin \left(2A\right)=2\sin \left(A\right)\cos \left(A\right)$ therefore,

$\cos \left(2A\right)={\cos }^2\left(A\right)-{\sin }^2\left(A\right)={\left(\frac{15}{17}\right)}^2-{\left(\frac{-8}{17}\right)}^2=\frac{225}{289}-\frac{64}{289}=\frac{161}{289}$

$\sin \left(2A\right)=2\sin \left(A\right)\cos \left(A\right)=2\times \frac{15}{17}\times \frac{-8}{17}=\frac{-240}{289}$

Hence,

$\cos (2A+B)=\cos \left(2A\right)\cos \left(B\right)-\sin \left(2A\right)\sin \left(B\right)=(\frac{161}{289}\times \frac{-24}{25})-(\frac{-240}{289}\times \frac{-7}{25})=\frac{-5544}{7225}$