Question

Assume cos (A-B) = cos A cos B + sin A sin B , find cos 15° when A = 45°, B = 30°

Gyus plz answer my question ​

Answer 1

Answer:

$\dfrac{ \sqrt{3} + 1 }{2 \sqrt{2} }$

Step-by-step explanation:

Just substitute the values for A and B in:

$\\\tt \cos \: A \cos \: B + \sin \: A\sin \: B \\ \\\tt = \cos 45 \cos 30 + \sin45\sin 30 \\ \\\tt= (\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} ) + ( \frac{1}{ \sqrt{2} } \times \frac{1}{2} ) \\ = \dfrac{ \sqrt{3} }{2 \sqrt{2} } + \dfrac{1}{2 \sqrt{2} } \\ = \dfrac{ \sqrt{3} + 1 }{2 \sqrt{2} }$

If you want you can rationalise it furthur.

Answer 2

Given :-

cos (A-B) = cos A cos B + sin A sin B ,

A = 45°, B = 30°

To Find :-

Cos 15°

Solution :-

We know that

$\sf cos \; 45^\circ = \dfrac{1}{\sqrt{2}}$

$\sf cos \; 30^\circ = \dfrac{\sqrt{3}}{2}$

$\sf sin \; 45^\circ = \dfrac{1}{\sqrt{2}}$

$\sf sin \; 30^\circ = \dfrac{\sqrt{3}}{2}$

$\sf \bigg(\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2}\bigg) + \bigg(\dfrac{1}{\sqrt{2}}\times\dfrac{1}2\bigg)$

$\sf \dfrac{\sqrt{3}}{\sqrt{2}\times 2} + \dfrac{1}{2 \times \sqrt{2}}$

$\sf \dfrac{\sqrt{3}+1}{\sqrt{2}\times 2}$

$\sf \dfrac{\sqrt{3}+1}{ 2\sqrt{2}}$