Question

Given that cos(A -B) = cod A.cosB + sinA.sinB, find the value of cos 15° in two ways.
(a) Taking A = 60° , B = 45°
(b) Taking A = 45° , B = 30°

Step by step explanation
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Answer 1

Given ,

$\sf {\green{cos(A-B)=cosA.cosA+sinA.sinB}}$

( a ) Taking   A = 60° , B = 45°

$\bf cos(60-45)\\\\ \implies \bf cos60.cos45+sin60.sin45 \\\\\implies \bf \dfrac{1}{2}.\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}.\dfrac{1}{\sqrt{2}}\\\\\implies \bf{\red{ \dfrac{\sqrt{3}+1}{2\sqrt{2}}}}$

( b ) Taking   A = 45° , B = 30°

$\bf cos(45-30)\\\\\implies \bf cos45.cos30+sin45.sin30\\\\\implies \bf \dfrac{1}{\sqrt{2}}.\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}.\dfrac{1}{2}\\\\\implies \bf {\blue{\dfrac{\sqrt{3}+1}{2\sqrt{2}}}}$