It is proved that cos ^ a-SiN ^ a = 1-2sin ^ a = 2cos ^ A-1 = cos ^ a-SiN ^ a

It is proved that cos ^ a-SiN ^ a = 1-2sin ^ a = 2cos ^ A-1 = cos ^ a-SiN ^ a

According to the cosine 2-fold angle formula cos (a + b) = cosa * cosb-sina * sinbcos2a = cos (a + a) = cosa * cosa Sina * Sina = cos? A-SiN? A, and then according to the identity of trigonometric function sin? 2A + cos? A = 1, cos? A-SiN? A = 1-sin? A-SiN? Sup

It is proved that sin ^ 4 α - cos ^ 4 α = 2Sin α 2 - 1

Left = (sin? α + cos? α) (sin? α - cos? α)
=sin²α-cos²α
=sin²α-(1-sin²α)
=2sin²α -1
=Right

It is proved that (1-2sin θ cos θ) / (COS ^ 2 θ - Sin ^ 2 θ) = (COS ^ 2 θ - Sin ^ 2 θ) / (1-2sin θ cos θ)

prove:
Left = 2Sin (П + θ) cos θ - 1 / 1-2sin ^ 2 θ
=(-2sinθcosθ-1)/cos2θ
=-(2sinθcosθ+sin^2 θ+cos^2 θ)/(cos^2 θ-sin^2 θ)
=-(sinθ+cosθ)^2/(cosθ-sinθ)(cosθ+sinθ)
=-(sinθ+cosθ)/(cosθ-sinθ)
=-[(sinθ/cosθ)+1]/[1-(sinθ/cosθ)]
=-(tanθ+1)/(1-tanθ)
=(tanθ+1)/(tanθ-1)
Right = Tan (9 П + θ) - 1 / Tan (П + θ) + 1
=(tanθ-1)/(tanθ+1)

If sin θ - 2cos θ = 0, find: (1) the value of sin ^ 2 θ + sin θ cos θ - 2cos ^ θ, (2) the value of 1 / cos ^ 2 θ + 2Sin θ cos θ

(1) the original formula = (sin? 2x + sinxcosx-2cos? X) / (sin? 2x + cos? X) = (tan? X + tanx-2) / (tan? X + 1) = 4 / 5. (2) the original formula = (sin? X + cos? X) / (COS? X + 2sinxcosx) = (tan

Cos α * (2cos β - cos α) + sin α * (2Sin β - sin α) = 0, calculate the value of β - α

cosα*(2cosβ-cosα)+sinα*(2sinβ-sinα)=0
2cosαcosβ-cos^2α+2sinαsinβ-sin^2α=0
2(cosαcosβ+2sinαsinβ)-1=0
cos(α-β)=1/2
cos(β-α)=1/2
β-α=2kπ±π/3（k∈Z）

If cos2a = 1 / 3, then the value of SiN4 power a + cos 4 power a is

It is known that cos2a = 1 / 3
Because cos2a = 2cos2 power A-1 = 1-2sin2 power a
So Cos2 power a = 2 / 3, sin 2 power a = 1 / 3
Sin 4 power a + cos 4 power a
=(sin2 power a + Cos2 power a) ^ 2-2sin2 power a * Cos2 power a
=1-2*(1/3)*(2/3)
=5/9
Then the value of sin 4 power a + cos 4 power a is 5 / 9

If cos2a = 1 / 4, then the value of Cos4 power a + SiN4 power a + sin? ACOS? A is equal to?

Cos 4 power a + sin 4 power a + sin 2 a = cos 4 power a + 2 sin 2 ACOS 2 A + sin 4 A - Sin 2 ACOS 2 a = (sin 2 A + cos 2 a) 2 - Sin 2 ACOS 2 a = 1 - 1 / 4 sin 2 a = 1 - 1 / 4 (1 - cos 2 a) = 49 / 64

sin6°+cos15°sin9°/cos6°-sin15°sin9°

Sin6 ° + cos15 ° sin9 ° = sin (15-9) + cos15sin9 = sin15cos9-sin9cos15 + cos15sin9 = sin15cos9cos6 ° - sin15 ° sin9 = cos (15-9) - sin15sin9 = sin15sin9-cos15cos9-sin15cos9 = - cos15cos9 so the original formula = sin15 / (- cos15) = - tan15 = √ 3-2

Calculation: (sin9 ° + cos15 ° sin6 °) / (cos9 ° - sin15 ° sin6 °)

(sin9 ° + cos15 ° sin6 °) / (cos9 ° - sin15 ° sin6 °) = (sin (15-6) + cos15 * sin6) / (COS (15-6) - sin15 * sin6) = (sin15 * cos6-cos15 * sin6 + cos15 * sin6) / (cos15 * cos6 + sin15 * sin6) = sin15 * cos6 / (cos15 * cos6) = tan15 = 2-root 3 (...)

(cos15°sin9°+sin6°)/(sin15°sin9°-cos6°)=

The original formula = [cos15 ° sin9 ° + sin (15 ° - 9)] / [sin15 ° sin9 cos (15 ° - 9 °)] = (cos15 ° sin9 + sin15 ° cos9 ° - cos15 ° sin9 °) / (sin15 ° sin9 ° - cos15 ° cos9 °) = sin15 ° cos9 ° (- cos15 ° cos9 °) = sin15 ° / cos15 ° = - tan15