Question

Sin^2x-cos^2=Корень2/2

Answer 1

Применим формулу:
cos2α = cos∧2x - sin∧2x; -( cos∧2x - sin∧x) =  - cos2α
- cos(2α) = √2/2
cos(2α) = - √2/2
2α = (+ -) arccos (√2/2) + 2πn, n∈Z
2α =( + -) (π -  π/4) + 2πn,n∈Z
2α = (+ -) (3π/4) + 2πn. n∈Z
α =( + -) (3π/8) +  πn, n∈Z