Question

if A= 45° verify that cos 2A= (2 cos^2 A -1) = 1-2 sin^2A​

Answer 1

Proof

Here given A=45°

Now Let me write some values here so that we would not have to look in books for these values

cos 45°=sin 45°=1/√2

and cos 90°=0

so now LHS = cos 2A

=cos 2×45°

= cos 90°

=0

now MHS = (2cos^2A-1)

= {2×(1/√2)^2-1}

= (2×1/2-1)

= 1-1

=0

now RHS = (1-2sin^2A)

={1-2(1/√2)^2}

=1-(2×1/2)

= 1-1

=0

So LHS = MHS = RHS