y=(sin x + COS x)^2

y=(sin x + COS x)^2

y=1+sin2x

Cos [π / 2 - π X / 2] = sin π X / 2

cos(π/2)cosπx/2+sinπ/2sinπx/2
=0+sinπx/2
=sinπx/2

: how does cos x represent SiN x / 2, cos X / 2, Tan X / 2? How should sin x, cos x represent Tan X / 2

Cosine formula from double angle: cos 2x = 1-2 (SiN x) ^ 2 = 2 (COS x) ^ 2-1
Because X and X / 2 are also double, so: cos x = 1-2 (SiN x / 2) ^ 2 = 2 (CoS / 2) ^ 2 - 1
Under the simplification, we get: SiN x / 2 = under the root sign (1-cos x) / 2
Cos X / 2 = under (1 + cos x) / 2
tan x=sin x/cos x
tan x/2=（sin x/2）/（cos x/2）
Just put in what you just said
So: Tan X / 2 = [1-cos X / 2] / [cos X / 2 = 1 + cos X / 2]
=(1-cos x) / (1 + cos x) under radical

Why sin (half π - α) = cos α?

In the right triangle ABC, a, B and C are the opposite sides of ∠ a, B and C, respectively, ∠ C = 90 degree
sinA=a/c
cosB=a/c
sinA = cosB = a/c
B = π / 2 - A
∴sinA = cos(π/2-A) = a/c

Simplification √ 1-sin8 + √ 2 + 2cos8

In the first root, 1-sin8 = 1-2 · SiN4 · Cos4 = sin4-cos4.4 ＞ 5 π / 4, so the first root is reduced to cos4-sin4; in the second root, 2 + 2cos8 = 4 (Cos4) ^ 2, so the second root is reduced to - 2cos4, that is

The results of 2 + 2cos8 + 21 − sin8 are as follows______ ．

∵ π < 4 < 3 π 2, ∵ Cos4 < 0, sin4-cos4 < 0, then the original formula = 2 + 2 (2cos24 − 1) + 2 (SiN4 − Cos4) 2 = 2 | Cos4 | + 2 | sin4-cos4 | = - 2cos4 + 2cos4-2sin4 = - 2sin4

Simplifying tan2 θ - Tan θ / 1 + tan2 θ Tan θ

(tan2θ-tanθ)/(1+tan2θtanθ)
=Tan (2 θ - θ)
= tanθ

(Tan α, Tan 2 α) / (Tan 2 α - Tan α) reduction

Because Tan α = Tan (2 α - α) = (tan2 α - Tan α) / (1 + tan2 α, Tan α)
So (Tan α, Tan 2 α) / (Tan 2 α - Tan α) = (Tan α, Tan 2 α) / [(1 + Tan 2 α, Tan α) * Tan α] = Tan 2 α / (1 + Tan 2 α, Tan α)
=sin2αcosα/(cos2αcosα+sin2αsinα)=sin2αcosα/cosα=sin2α

Simplifying Tan θ · Tan 2 θ + Tan 2 θ · Tan 3 θ +. + Tan · θ * Tan (n + 1) θ

∵1+tannθ*tan(n+1)θ=[cosnθcos(n+1)θ+sinnθsin(n+1)θ]/cosnθcos(n+1)θ
=cos[(n+1)θ-nθ]/cosnθcos(n+1)θ
=cosθ/cosnθcos(n+1)θ
=cotθ*sinθ/cosnθcos(n+1)θ
=cotθ*sin[(n+1)θ-nθ]/cosnθcos(n+1)θ
=cotθ[sin(n+1)θcosnθ/cosnθcos(n+1)θ-cos(n+1)θsinnθ/cosnθcos(n+1)θ]
=cotθ*[tan(n+1)θ-tannθ]
∴tannθ*tan(n+1)θ=cotθ*[tan(n+1)θ-tannθ]-1
So tan θ · Tan 2 θ + Tan 2 θ · Tan 3 θ +. + Tan · θ * Tan (n + 1) θ
=[cotθ(tan2θ-tanθ)-1]+[cotθ(tan3θ-tan2θ)-1]+.+cotθ[tan(n+1)θ-tannθ]-1
=cotθ[tan(n+1)θ-tanθ]-n
=cotθtan(n+1)θ-n-1
=tan(n+1)θ/tanθ-n-1

Y + (Tan α) ^ 2 + 2tan α reduction

Original formula = (Tana + 1) ^ 2 + Y-1