Given TaNx = 3, find the value of sinxcosx

Given TaNx = 3, find the value of sinxcosx

sinxcosx=sinxcosx/(sin^2x+cos^2x)=tanx/(1+tan^2x)=3/10

If f (TaNx) = cos2x, then f (2) is

Answer: let cosx = t, then TaNx = SiNx / cosx = √ (1-T ^ 2) / T, cos2x = 2 (cosx) ^ 2-1 = 2T ^ 2-1, so f (√ (1-T ^ 2) / T) = 2T ^ 2-1. Let √ (1-T ^ 2) / T = u, then t = 1 / √ (u ^ 2 + 1), substituting t = 1 / √ (u ^ 2 + 1) into f (U) = 2T ^ 2-1, there is f (U) = 2 / (u ^ 2 + 1) - 1, that is, f (x) = 2 / (x ^ 2 + 1) - 1

The minimum positive period of function f (x) = sin2x * 1 / TaNx

f(x)=sin2x*1/tanx
=2sinxcosx*1/tanx
=2sinxcosx*1/[sinx/cosx]
=2sinxcosx* cosx/sinx
=2cos²x
=1+cos(2x),
So the period of the function is 2 π / 2 = π