Question

1: cos 18° - sin 18° = root 2sin 27° ​

Answer 1

Answer:

It has been proved that:

$cos18\°-sin18\°=\sqrt 2\,sin27\°$

Step-by-step explanation:

L.H.S

$cos18\°-sin18\°$

$=cos18\°-cos(90\°-72\°)$

$=cos18\°-cos72\°$

Here we are using the formula of $cosC-cosD$.

$cosC-cosD=-2sin(\dfrac{C+D}{2})sin(\dfrac{C-D}{2})$

$cos18\°-cos72\°=-2sin(\dfrac{18+72}{2})sin(\dfrac{18-72}{2})$

                               $=-2sin45\°\,sin(-27\°)$

As we know:

$sin(-x)=-sinx$

Therefore:

$=-2\cdot \dfrac{1}{\sqrt 2}(-sin27\°)$

$=\dfrac{\sqrt 2\codt \sqrt 2}{\sqrt 2}sin(27\°)$

$=\sqrt 2 sin(27\°)$

=R.H.S