Question

If $\sin \left({40}^{\circ }+\theta \right)=b,0<\theta <{45}^{\circ }$, find

(a) $\cos \left({70}^{\circ }+\theta \right)$
(b) $\cos \left({160}^{\circ }+\theta \right)$

Answer 1

Given that,

$\sin ({40}^0+\theta )=b$

Then,

$\cos ({40}^0+\theta )=\sqrt{1-{\sin }^2({40}^0-\theta )}$

$=\sqrt{1-b^2}$

Part (a),

$\cos ({70}^0+\theta )=\cos [{30}^0+({40}^0+\theta )]$

$=\cos {30}^0\cos ({40}^0+\theta )-\sin {30}^0\sin ({40}^0+\theta )$

$=\frac{\sqrt{3}}{2}\sqrt{1-b^2}-\frac{1}{2}b$

$=\frac{\sqrt{3}\sqrt{1-b^2}-b}{2}$

Hence, this is the answer.

Part(b)

$\cos ({160}^0+\theta )=\cos [{120}^0+({40}^0+\theta )]$

$=\cos {120}^0\cos ({140}^0+\theta )-\sin {120}^0\sin ({140}^0+\theta )$

$=\cos ({180}^0-{60}^0)\cos ({140}^0+\theta )-\sin ({180}^0-{60}^0)\sin ({140}^0+\theta )$

$=-\cos {60}^0\sqrt{1-b^2}-\sin {60}^{0}b$

$=-\frac{1}{2}\sqrt{1-b^2}-\frac{\sqrt{3}}{2}b$

$=-\left(\frac{\sqrt{1-b^2}+\sqrt{3}b}{2}\right)$

Hence, this is the answer.