Simplification: [cos (α - π / 2) / sin (5 π / 2 + α)] × sin (α - 2 π) × cos (2 π - α) RT.

Simplification: [cos (α - π / 2) / sin (5 π / 2 + α)] × sin (α - 2 π) × cos (2 π - α) RT.

The original formula = sin α / cos α * sin α * cos α
=sin^2α

Simplification of sin (θ - 5 π) cos (π / 2 + θ) cos (4 π - θ) / cos (3 π - θ) sin (θ - 3 π) sin (- θ - 4 π)

The original formula = - sin (θ) * - 1) sin (θ) cos (θ) / - cos (θ) * - 1 sin (θ) * - 1 sin (θ) = - 1

Simplification: sin ^ 3 (π / 2 + a) + cos ^ 3 (3 π / 2-A) / sin (3 π + a) + cos (4 π - a) - sin (5 π / 2 + a) * cos (3 π / 2 + a)

=(cos^3a-sin^3a)/(-sina+cosa)-cosasina
=(cosa-sina)(cos^2a+cosasina+sin^2a)/(cosa-sina)-cosasina
=cos^2a+cosasina+sin^2a-cosasina
=cos^2a+sin^2a
=1

Confirmation: tana-1 / Tana =) 1-2cos? A) / (Sina * COSA)

tana-1/tana=(tan^2a-1)/tana=cos^2a(tan^2a-1)/(tanacos^2a)
=(sin^2a-cos^2a)/(sinacosa)
=(1-cos^2a-cos^2a)/(sinacosa)
=(1-2cos^2a)/(sinacosa)

The maximum value of X (2x) sin (2) is obtained

(x) + 3sin (x) cos (x) - 2cos ＾ 2 (x) = cos ^ 2 (x) + sin ^ 2 (x) + 3sin (x) cos (x) + 3sin (x) cos (x) - 3cos ^ 2 (x) = 1 + 3sin (x) cos (x) - 3cos ^ 2 (x) = 1 + 3sin (x) cos (x) - 3cos ^ 2 (x) = 1 + 3 / 2Sin (2x) - 3 / 2 [cos (2x) + 1] = 3 / 2Sin (2x) - 3 / 2cos (2x) - 1 / 2 = 1 / 2 = 3 / root number (root) 3 / root number (2x) - 3 / 2cos (2x) - 1 / 2 = 3 / 3 / root root number (root) 3 / root number (root number) - 3 / root number (root) root 2 [sin (2x) cos π / 4-cos (2x) sin π / 4] - 1

If cos α + 3sin α / 6sin α - 2cos α = 2 (1) find Tan α (2) calculate sin2 ^ α + 3sin α cos α - 2cos2 ^ α

(COS α + 3sin α) / (6sin α - 2cos α) = 2cos α + 3sin α = (6sin α - 2cos α) * 25cos α = 9sin α Tan α = 5 / 9 sin2 ^ α + Cos2 ^ α = 1, so cos α = 9 / (106) ^ 0.5 sin α = 5 / (106) ^ 0.5sin2 ^ α + 3sin α cos α - 2cos2 ^ α = - 1 / 53

Given that 3sin α - 2cos α = 0, calculate the value of (COS α - sin α) / (COS α + sin α) + (COS α + sin α) / (COS α - sin α)

Because cos α = 1.5sin α, so
Original formula = 0.5 / 2.5 + 2.5 / 0.5 = 1 / 5 + 5 = 5.2

Sin α = - 2cos α, find sin ^ 2 α - 3sin α cos α + 1

sina=-2cosatana=-2sin²a-3sinacosa+1=(sin²a-3sinacosa+sin²a+cos²a)/(sin²a+cos²a)=(tan²a-3tana+tan²a+1)/(tan²a+1)=(4+6+4+1)/(4+1)=3

Given that 3sin α - 2cos α = 0, calculate the value of (COS α - sin α) / (COS α + sin α) Given that 3sin α - 2cos α = 0, calculate the values of the following equations ⑴((cosα-sinα)/(cosα+sinα))+((cosα+sinα)/(cosα-sinα))= ⑵sin²α-2sinαcosα+4cos²α=

The first question answer is 26 / 5, the second question is 28 / 13. The solution is Sina = (2 / 3) cosa, and then replace Sina with cosa. The second problem is to divide the whole equation by (COSA) square + (Sina) square, and then replace Sina with cosa

∫ arctan root (x) / root (x) * (1 + x) Let t = radical X x=t^2 ∫arctan t/t(1+t^2)dt^2 How to continue?

Analysis: let t = √ x, then x = t?, DX = DT? = 2tdt, so the original formula = ∫ (arctan √ x) / √ x (1 + x) DX = ∫ [arctant / T (1 + T?)] * 2tdt = 2 ∫ arctant / (1 + t?) DT = 2 ∫ arctant D (arctant) = 2 * 1 / 2 * (arctant) 2 + C. = (arc