Why? F (SiNx) = cos2x = 1-2 (SiNx) ^ 2 Why does f (SiNx) = cos2x have 1-2 (SiNx) ^ 2 equal to them

Why? F (SiNx) = cos2x = 1-2 (SiNx) ^ 2 Why does f (SiNx) = cos2x have 1-2 (SiNx) ^ 2 equal to them

cos2x=cos(x+x)
=cosxcosx-sinxsinx
=1-(sinx)^2-(sinx)^2
=1-2(sinx)^2

Tan (π / 4-A) = 1 / 3, the value of Sin & sup2; a-2cos & sup2; a is obtained

tan(π/4-a)=tan(π/4)-tan(a)/1+tan(π/4)tan(a)
So tan (a) = 1 / 2
Because Sin & sup2; a-2cos & sup2; a = Sin & sup2; a-2cos & sup2; a / Sin & sup2; a + cos & sup2; a
At the same time divided by cos & sup2; a
=-7/5
By the way, how do you type cos & sup2; a

It is known that Tan α = √ 2 can be used to find the value of 2cos & sup2; α / 2-sin α - 1 / sin α + cos α

Let the three sides of a right triangle be a B C, Tan α = A / b = √ 2, that is, a = √ 2B, then the hypotenuse C = √ 3b, so sin α = A / C = √ 2 / √ 3, cos α = B / C = 1 / √ 3 2cos & # 178; α / 2-sin α - 1 / sin α + cos α = 2 × (1 / 3) / 2 - √ 2 / √ 3-1 / (√ 2 / √ 3) + 1 / √ 3 = (2-5 √ 6 +