Question

if A=60 degree and B=30 degree, verify that

cos (A-B)=cosAcosB + sinAsinB

Answer 1

A=60° B=30°

Cos(60°-30°)=cos60°×cos30°+sin60°×sin30°
cos30°=1/2×√3/2+√3/2×1/2
cos30°=√3/4+√3/4
cos30°=2√3/4
cos30°=√3/2
LHS. = RHS

Since cos30°=√3/2
Hence verified
Plzzz mark as brainlist yar

Answer 2

Answer:

Given information: $A=60^{\circ}$ and $B=30^{\circ}$.

We need to verify the formula

$\cos (A-B)=\cos A\cos B+\sin A \sin B$

Taking LHS of the given formula.

$LHS=\cos (A-B)$

Substitute the given values.

$LHS=\cos (60^{\circ}-30^{\circ})$

$LHS=\cos (30^{\circ})$

$LHS=\frac{\sqrt{3}}{2}$

Taking RHS of the given formula.

$RHS=\cos A\cos B+\sin A \sin B$

Substitute the given values.

$RHS=\cos 60^{\circ}\cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}$

$RHS=\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}$

$RHS=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$

$RHS=\frac{2\sqrt{3}}{4}$

$RHS=\frac{\sqrt{3}}{2}$

Hence proved LHS =RHS.