Question

If $\sin ({40}^o+\theta )=b$, $0<\theta <45$. find $\cos ({70}^o+\theta )$

Answer 1

$\sin \left({40}^{\circ }+\theta \right)=b------\left(1\right)$

$\left(1\right)-----\cos \left({70}^{\circ }+\theta \right)$

$=\cos \left({30}^{\circ }+{40}^{\circ }+\theta \right)$

$=\cos {30}^{\circ }\cos \left({40}^{\circ }+\theta \right)-\sin {30}^{\circ }\sin \left({40}^{\circ }+\theta \right)$

$=\frac{\sqrt{3}}{2}\sqrt{1-{\sin }^2\left({40}^{\circ }+\theta \right)}-\frac{1}{2}\times b$

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

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