Question

Given that cos (A-B) = Cos A CosB + sina sinb
find the value of Cos 15° In two ways
a) Taking A = 60°, B = 45°and
b) Taking A=45°, B = 30°​

Answer 1

Answer

cos 15 = ( √3 + 1 ) / 2√2

Given :

• Cos(A-B) = CosA.CosB + SinA.SinB

To Find :

• Value of Cos 15° In two ways

a) Taking A = 60°, B = 45°and

b) Taking A=45°, B = 30°​

Solution :

$\bf a)\ Cos(60-45)\\\\\implies \sf Cos60.Cos45+Sin60.Sin45\\\\\implies \sf \dfrac{1}{2}.\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}.\dfrac{1}{\sqrt{2}}\\\\\implies \sf \dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}\\\\\implies \bf \dfrac{\sqrt{3}+1}{2\sqrt{2}}$

$\bf b)\ Cos(45-30)\\\\\implies \sf Cos45.Cos30+Sin45.Sin30\\\\\implies \sf \dfrac{1}{\sqrt{2}}.\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}.\dfrac{1}{2}\\\\\implies \sf \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\\\\\implies \bf \dfrac{\sqrt{3}+1}{2\sqrt{2}}$