Question

Find the value of:

(i) $\sin {75}^o$

(ii) $\tan {15}^o$

Answer 1

i) $\sin {75}^o=\sin ({90}^o-{15}^o)=\cos {15}^o$ {$sin({90}^o-A)=\cos A$}

using identity $\cos \frac{A}{2}=\sqrt{\frac{1+\cos A}{2}}$

put $A={30}^o$ in above identity

Therefore, $\cos {15}^o=\sqrt{\frac{1+\cos {30}^o}{2}}=\frac{\sqrt{\sqrt{3}+2}}{2}$

ii) $\tan {15}^o=\sqrt{{\sec }^2{15}^o-1}=\sqrt{\frac{1}{{\cos }^2{15}^o}-1}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$ from(i)