1 + 2cos 2 Ø - Cos2 Ø

1 + 2cos 2 Ø - Cos2 Ø

1+2cos²Ø-cos2Ø
=1+1+cos2Ø-cos2Ø
=2
Reduction formula:
(cosX)^2=(1+cos2X)/2
(sinX)^2=(1-cos2X)/2

Simplification (1 + cos α + Cos2 α + cos3 α) / (2cos 2 α + cos α - 1)

The applied formula and difference product are as follows: cos α + cos β = 2cos (α + β) / 2cos (α - β) / 2 double angle cosine: 2cos 2 α - 1 = Cos2 α, 1 + Cos2 α = 2cos 2 α ν (1 + cos α + Cos2 α + cos3 α) / (2cos 2 α + cos α - 1) = [(1 + Cos2 α) + (COS α + cos3 α)] / (cos

What is the difference between Cos2 and COS 2 and 2cos? I don't know their relationship

Cos2a = 2cos square a minus 1

Simplification of sin α cos α Cos2 α

sinαcosαcos2α
=1/2sin2αcos2α
=1/4sin4α

Simplification: Cos2 α / (COS α - sin α) - sin α

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It is proved that for any angle θ,
∵cos4θ-sin4θ=（cos2θ+sin2θ）•（cos2θ-sin2θ）
=cos2θ-sin2θ=cos2θ，
The result shows that cos 4 θ - Sin 4 θ = cos 2 θ holds

Simplification of (sin2x + 2sinx ^ 2) / 1-tanx

① (sin2x + 2sinx?) / 1-tanx = 2sinx (cosx + SiNx) / [(cosx SiNx) / cosx] = 2sinxcosx (cosx + SiNx) 2 / (COS? X-sin? X) = tg2x · (cosx + SiNx) 2? Yes No;; 1-tanx should be 1 + TaNx, if it is (sin2x + 2S

Simplification of sin2x (1 + TaNx · TaNx) 2) The result is______ .

sin2x(1+tanx•tanx
2)=sin2x（1+sinxsinx
Two
cosxcosx
2）=sin2x（1+2sinx
2sinx
Two
cosx）=sin2x（1+1−cosx
cosx）=2sinx
So the answer is: 2sinx

Simplification of function f (x) = 1 / (1 + (TaNx) ^ 2) + (1 + cos2x) / 2

f(x)=1/(cosx^2+sinx^2)*cosx^2+2cosx^2/2=cosx^2+cosx^2=2cosx^2

2A vector = (sin2x, cos2x), B = (cos2x, - cos2x) 2. A vector = (Radix 3sin2x, cos2x), B = (cos2x, - cos2x) (1) X ∈ (7 / 24 Π, 5 / 12 Π), a * B + 1 / 2 = - 3 / 5, find cos4x (2) The three sides of the triangle ABC are a, B, C respectively, and B * b = AC, the angle corresponding to B side is x, and a vector * B vector + 1 / 2 has and only one real root

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