Question -
Answer -
(i) cos (3π/4 + x) – cos (3π/4 – x) = -√2 sin x
Let us consider LHS:
cos (3π/4 + x) – cos (3π/4 – x)
By using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
cos (3π/4 + x) – cos (3π/4 – x) = -2 sin (3π/4 + x + 3π/4 – x)/2 sin (3π/4 + x – 3π/4 + x)/2
= -2 sin (6π/4)/2 sin 2x/2
= -2 sin 6π/8 sin x
= -2 sin 3π/4 sin x
= -2 sin (π – π/4) sin x
= -2 sin π/4 sin x (since, (π-A) = sin A)
= -2 × 1/√2 × sin x
= -√2 sin x
= RHS
Hence proved.
(ii) cos (π/4 + x) + cos (π/4 – x) = √2 cos x
Let us consider LHS:
cos (π/4 + x) + cos (π/4 – x)
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos (π/4 + x) + cos (π/4 – x) = 2 cos (π/4 + x + π/4 – x)/2 cos (π/4 + x – π/4 + x)/2
= 2 cos (2π/4)/2 cos 2x/2
= 2 cos 2π/8 cos x
= 2 sin π/4 cos x
= 2 × 1/√2 × cos x
= √2 cos x
= RHS
Hence proved.