Решите пожалуйста выражение!
Cos2α=cos²α-sin²α 1)√2cos²(3π/8)-√2sin²(3π/8)=√2(cos²(3π/8)-sin²(3π/8))= =√2cos(6π/8)=√2cos(3π/4)=√2(-√2/2)=-1 2)√3cos²(π/12)-√3sin²(π/12)=√3(cos²(π/12)-sin²(π/12))= =√3cos(2π/12)=√3cos(π/6)=√3√3/2=3/2=1,5