Question

If $({60}^{\circ }+\theta )$ and $({30}^{\circ }-\theta )$ are acute angles, find the value of $sin({60}^{\circ }+\theta )-cos({30}^{\circ }-\theta )$.

Answer 1

Given that,
$({60}^{\circ }+\theta )$ and $({30}^{\circ }-\theta )$ are acute angles.
We know, $(60+\theta )$ + $(30-\theta )$ = 90°
Therefore, $(60+\theta )$ = 90° - $(30-\theta )$

$sin({60}^{\circ }+\theta )$ - $cos({30}^{\circ }-\theta )$
$=sin[{90}^{\circ }-({30}^{\circ }-\theta \left)\right]-cos({30}^{\circ }-\theta )$

$=cos({30}^{\circ }-\theta )-cos({30}^{\circ }-\theta )$
= 0