Equation of tangent line

Question

f(x)=ln√1+sin²x (pi/4,ln√3/2)

Answer 1

Best to rewrite f(x) as 1/2 [ln(1+sin²x)].

Then f'(x) = 1/2 (1/(1+sin²x)) (2sinx cosx). = sinxcosx/(1+sin²x)

Find the slope of the tangent line at π/4, as f'(π/4) = (√2/2)(√2/2)/(3/2) = 1/3.

Using point slope: y - ln(√(3/2)) = 1/3 (x - π/4).

Note that f(π/4) is indeed ln(√(3/2))