Question

Given that : cos(A-B) =cosAcosB+sinAsinB find the value of cos 15 degree .
please help me solve the problem. the first and with good explanation answer will get the Brainliest answer

Answer 1

HEYA!!
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Cos 15 ° = Cos ( 45° - 30 ° )

Cos ( A- B ) = CosA CosB + SinA SinB.

Cos( 45° - 30° ) = Cos45. Cos30+ Sin45.Sin30

=
$\frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } . \frac{1}{2}$
$\frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} } \\ \\ = \frac{ \sqrt{3} + 1 }{2 \sqrt{2} } \times \frac{2 \sqrt{2} }{2 \sqrt{2} } \\ \\ = \frac{2 \sqrt{6} + 2 \sqrt{2} }{8} \\ \\ = \frac{2( \sqrt{6} + \sqrt{2} )}{8} \\ \\ = \frac{ \sqrt{6} + \sqrt{2} }{4}$
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Answer 2

Answer:

U have posted the question half. In given question I have posted the question and answer

I have posted the next half question