It is known that Tan α = 1 2, find 1 + 2Sin (π − α) cos (− 2 π − α) sin2(−α)−sin2(5π 2 − α)

It is known that Tan α = 1 2, find 1 + 2Sin (π − α) cos (− 2 π − α) sin2(−α)−sin2(5π 2 − α)

∵tanα=1
2，
The original formula = 1 + 2Sin α cos (2 π + α)
sin2α−sin2(π
2−α)
=1+2sinαcosα
sin2α−cos2α
=(sinα+cosα)2
(sinα+cosα)(sinα−cosα)
=sinα+cosα
sinα−cosα
=1+tanα
tanα−1
=1+1
Two
One
2−1
=-3．

It was proved that 2Sin (π / 2-A) * Tan (π - a) / (1 + cos2a) * cos (- a) = - Tana

2cosa -tana/2cos²a cosa=-tana

It is proved that the following identity Tan ^ 2 θ * (1-sin θ) / (1 + cos θ) = (1-cos θ) / (1 + sin θ) Detailed process

Immediate proof
tan^2 θ=[(1-cosθ)/(1+sinθ)]*[(1+cosθ)/(1-sinθ)]
=[(1-cosθ)(1+cosθ)]/[(1-sinθ)(1+sinθ)]
=(1-cos^2 θ)/(1-sin^2 θ)
=sin^2 θ/cos^2 θ
Obviously, it can be reversed

It is proved that the identity Tan α sin α / Tan α - sin α = 1 + cos α / sin α

tanαsinα/(tanα-sinα)=sinα/cosα*sinα/(sinα/cosα-sinα)=[(sinα)^2/cosα]/[(sinα-sinαcosα)/cosα]=[(sinα)^2/cosα]*cosα/(sinα-sinαcosα)=(sinα)^2/(sinα-sinαcosα)=sinα/(1-cosα)=sin...

It is proved that the following identity (sin θ + cos θ) / (1-tan ^ 2 θ) + sin ^ 2 θ / (sin θ - cos θ) = sin θ + cos θ

(sinθ+cosθ)/(1-tan²θ)+sin²θ/(sinθ-cosθ)=(sinθ+cosθ)/(1-sin²θ/cos²θ)+sin²θ/(sinθ-cosθ)=(sinθ+cosθ)/[(cos²θ-sin²θ)/cos²θ]+sin²θ/(sinθ-cos...

It is proved that the identity (COS α + Tan α) / [(COS α / sin α) + 1 / cos α] = sin α It is proved that the identity (COS α + Tan α) / [(COS α / sin α) + 1 / cos α] = sin α

(cosα+tanα)/[(cosα/sinα)+1/cosα]= (cosα+sinα/cosα)/[(cosα/sinα)+1/cosα]=sinα{[(cosα/sinα)+1/cosα]/[(cosα/sinα)+1/cosα]}=sinα

Prove the identity: (1 + sin α) / cos α = (1 + Tan (α / 2)) / (1-tan (α / 2)) There is also a simplification: (COS (θ + 15 °) ^ 2 + (sin (θ - 15 °) ^ 2 + cos (θ + 180 °) sin (θ + 180 °)

1. For this kind of problem, we can draw a conclusion step by step. For this problem, we usually start from the side of Tan and cot. therefore, I choose from the right side to the left side. Tan (α / 2) = sin (α / 2) / cos (α / 2), It is simplified to get [cos (α / 2) + sin (α / 2)] / [cos (α / 2) - sin (α / 2)] / [cos (α / 2) - sin (α / 2)] = [1 + 2Sin (α / 2) cos (α / 2) cos (α / 2)] / [cos ^ 2 (α / 2) (α / 2) - Sin ^ 2 (α / 2) (α / 2)] {where the denominator is the denominator, the molecule is multiplied by cos (α / 2) + sin (α / 2)}, and then 2Sin (α / 2) cos (α / 2) cos (α / 2) = sin α, cos ^ 2 (α / 2) - Sin ^ 2 (α / 2) - Sin ^ 2 (α / 2) = cos α α α α α α α α, 2 (α / 2) - Sin ^ 2 (α / 2) = cos then we can get the certificate

It is proved that 1 / cos ^ α - Tan ^ 2 α - Sin ^ 2 α = cos ^ 2 α

1/cos²α -tan²α-sin²α
=1/cos²α - sin²α/cos²α -sin²α
=(1-sin²α)/cos²α -sin²α
=cos²α/cos²α -sin²α
=1-sin²α
=cos²α
The equation is proved

It is proved that Tan A / 2 = sin a / (1 + cos a)

Right side (2) cos (2) a / 2
=Sin (A / 2) / cos (A / 2) = left

The results showed that: (1 + sin α) / cos α = (1 + Tan α / 2) / (1-tan α / 2)

It is proved that: Sina = sin (A / 2 * 2) = s * sin (A / 2 * 2) = s * sin (A / 2 * 2) cosa = cos (A / 2 * 2) = (COS (A / 2 / 2)) ^ 2 (sin (a / 2)) ^ 2 (sin (A / 2)) ^ 2 puts Sina and cosa into (1 + sin α) / cos α = (sin (A / 2) + cos (A / 2)) ^ 2 / (sin (A / 2) - cos (A / 2)) / (sin (A / 2) + cos (A / 2)) = (sin (A / 2) + cos (A / 2)) = (sin (A / 2) + cos (A / 2)) / / (2) + cos (A / 2)) / (2)) / (2)) / (2)) / (2) + cos (A / 2 si