Given sin Χ + 2cos Χ = 0, find: Cos2 Χ 2 is the root of the sign, which is the root 2

Given sin Χ + 2cos Χ = 0, find: Cos2 Χ 2 is the root of the sign, which is the root 2

Given the formula, the formula with the sum of squares of 1 can calculate SiNx ^ 2 (or cosx ^ 2), and the angle cosine formula cos2x = 2cosx ^ 2-1 = 1-2sinx ^ 2 = cosx ^ 2-sinx ^ 2. There are three. You can use any one of them

It is proved that the identity Cos2 α / cos α - sin α = cos α + sin α

Cos (2 α) / (COS α - sin α) = (COS & # 178; α - Sin & # 178; α) / (COS α - sin α) / molecule uses the double angle formula = (COS α + sin α) (COS α - sin α) / (COS α - sin α) / molecule uses the square difference formula = cos α + sin α

The known sin angle / 2 + cos angle / 2 to find Cos2 multiple angle

As your question shows: (SiNx) / 2 + (cosx) / 2 = M
[(sinx)/2+(cosx)/2]^2=m^2
1/4+(sin2x)/4=m^2
So sin2x = (m ^ 2-1 / 4) times 4
=4m^2-1
Cos2x = [square of 1-sin2x]