Question -
Answer -
(i)┬атИЪ3 sin x тАУ cos x
Let f(x) = тИЪ3 sin x тАУ cos x
Dividing and multiplying by тИЪ((тИЪ3)2┬а+12) i.e. by 2
f(x) = 2(тИЪ3/2 sin x тАУ 1/2 cos x)
Sine expression:
f(x) = 2(cos ╧А/6 sin x тАУ sin ╧А/6 cos x) (since, тИЪ3/2 =cos ╧А/6 and 1/2 = sin ╧А/6)
We know that, sin A cos B тАУ cos A sin B = sin (A тАУ B)
f(x) = 2 sin (x тАУ ╧А/6)
Again,
f(x) = 2(тИЪ3/2 sin x тАУ 1/2 cos x)
Cosine expression:
f(x) = 2(sin ╧А/3 sin x тАУ cos ╧А/3 cos x)
We know that, cos A cos B тАУ sin A sin B = cos (A + B)
f(x) = -2 cos(╧А/3 + x)
(ii)┬аcos x тАУ sin x
Let f(x) = cos x тАУ sin x
Dividing and multiplying by тИЪ(12┬а+ 12)i.e. by тИЪ2,
f(x) = тИЪ2(1/тИЪ2 cos x тАУ 1/тИЪ2 sin x)
Sine expression:
f(x) = тИЪ2(sin ╧А/4 cos x тАУ cos ╧А/4 sin x) (since, 1/тИЪ2= sin ╧А/4 and 1/тИЪ2 = cos ╧А/4)
We know that sin A cos B тАУ cos A sin B = sin (A тАУ B)
f(x) = тИЪ2 sin (╧А/4 тАУ x)
Again,
f(x) = тИЪ2(1/тИЪ2 cos x тАУ 1/тИЪ2 sin x)
Cosine expression:
f(x) = 2(cos ╧А/4 cos x тАУ sin ╧А/4 sin x)
We know that cos A cos B тАУ sin A sin B = cos (A + B)
f(x) = тИЪ2 cos (╧А/4 + x)
(iii)┬а24 cos x + 7 sin x
Let f(x) = 24 cos x + 7 sin x
Dividing and multiplying by тИЪ((тИЪ24)2┬а+72) = тИЪ625 i.e. by 25,
f(x) = 25(24/25 cos x + 7/25 sin x)
Sine expression:
f(x) = 25(sin ╬▒ cos x + cos ╬▒ sin x)┬аwhere, sin ╬▒= 24/25 and cos ╬▒ = 7/25
We know that sin A cos B + cos A sin B = sin (A + B)
f(x) = 25 sin (╬▒ + x)
Cosine expression:
f(x) = 25(cos ╬▒ cos x + sin ╬▒ sin x)┬аwhere, cos ╬▒= 24/25 and sin ╬▒ = 7/25
We know that cos A cos B + sin A sin B = cos (A тАУ B)
f(x) = 25 cos (╬▒ тАУ x)