derivative of sin theta cosine theta?

Answer 1

lets rewrite this as

f(x) = sinxcosx
we use the product rule for this one.

first term: sinx
second term: cosx

product rule: derivative of the first term times the second + derivative of the second term times the first.

f'(x) = cosx(cosx) + -sinx(sinx)
f'(x) = cos^2x - sin^2x

Answer 2

although it should be made very clear that both the sine & cosine functions deal with different aspects of the fundamental functions in trignometry, they indeed are closely related & thus can be inter converted if necessary. first there is a relation between square of their values which equals 1. i.e. sin^2 x + cos^ 2x = 1 besides that we did have another relation. sin (90-x) = cos x & its opposite is also true i.e. cos (90-x) = sin x for example sin 60 = sqrt 3/2 now, sin (90 - 60) = cos 30 = sqrt 3/2 also, cos (90-30) = sin 60 = sqrt 3/2 or sin 30 = cos (90-30) = cos 60 = 1/2 etc.

Answer 3

let f(x)=sinxcosx
we know that sin2x=2sinxcosx
so f(x)=1/2*sin2x
f'(x)=1/2cos2x*2
=cos2x
hence cos2x is the answer

Answer 4

y = sin θ cos θ
dy/dθ = cos θ cos θ - sin θ sin θ
dy/dθ = cos ² θ - sin ² θ
dy/dθ = cos 2 θ