Question

If $\sin ({40}^{\circ }+\theta )=b,0<\theta <45,$, find
$i\left)\cos \right({70}^{\circ }+\theta )$
$ii\left)cos\right({160}^{\circ }+\theta )$

Answer 1

Given $\sin ({40}^{\circ }+\theta )=b,0<\theta <45$

$⟹\cos ({40}^{\circ }+\theta )=\sqrt{1-b^2}$

i)$\cos ({70}^{\circ }+\theta )=\cos ({30}^{\circ }+({40}^{\circ }+\theta \left)\right)=\cos {30}^{\circ }\cos ({40}^{\circ }+\theta )-\sin {30}^{\circ }\sin ({40}^{\circ }+\theta )=\frac{1}{2}(\sqrt{3-3b^2}+b)$

$⟹\sin ({70}^{\circ }+\theta )=\sqrt{1-\frac{1}{4}(\sqrt{3-3b^2}+b{)}^2}=\frac{1}{2}\sqrt{1+4b^2+2b\sqrt{3-3b^2}}$

ii)$\cos ({160}^{\circ }+\theta )=\cos ({90}^{\circ }+({70}^{\circ }+\theta \left)\right)=-\sin ({40}^{\circ }+\theta )=-\frac{1}{2}\sqrt{1+4b^2+2b\sqrt{3-3b^2}}$