Question

Express the following in terms of trigonometric ratios of angles lying between 0 and 45 :
$\sin {59}^0+\cos {56}^0$

Answer 1

We have,

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

$\sin {59}^0=\sin ({90}^0-{31}^0)=\cos {31}^0$

$\cos {56}^0=\cos ({90}^0-{34}^0)=\sin {34}^0$

$\therefore \sin {59}^0+\cos {56}^0=\cos {31}^0+\sin {34}^0$