Question

Express the following in terms of trigonometric ratios of angles lying between 0 and 45:
$\sin {67}^0+\cos {75}^0$

Answer 1

We have,

$\sin ({90}^0-\theta )=\cos \theta$ and $\cos ({90}^0-\theta )=\sin \theta$

$\sin {67}^0=\sin \left({90}^{-}{23}^0\right)=\cos {23}^0$

$\cos {75}^0=\cos ({90}^0-{15}^0)=\sin {15}^0$

$\therefore \sin {67}^0+\cos {75}^0=\cos {23}^0+\sin {15}^0$