If Tan α = 3, find the value of sin ^ 2 α - sin α cos α + 2cos ^ 2 α

If Tan α = 3, find the value of sin ^ 2 α - sin α cos α + 2cos ^ 2 α

The original formula = [sinacosa + 2 (COSA) ^ 2-2 (Sina) ^ 2] / [(Sina) ^ 2 + (COSA) ^ 2]
The same division of fractions (COSA) ^ 2,
Then the original formula = [Tana + 2-2 (Tana) ^ 2] / [(Tana) ^ 2 + 1] = - 13 / 10

Tan α = 1 / 2, find ① sin ^ 2 α + 2cos ^ 2 α, ② sin α * cos α + 2cos ^ 2 α

solution
sin²a+2cos²a
=(sin? 2A + 2cos? A) / (sin? 2A + cos? A) - divided by sin? 2A + cos? A = 1, the value remains unchanged
=(tan? A + 2) / (tan? A + 1) - the numerator and denominator are divided by cos? A at the same time
=(1/4+2)/(1/4+1)
=9/4×4/5
=9/5
sinacosa+2cos²a
=(sinacosa + 2cos? A) / (sin? 2A + cos? A)
=(tana+2)/(tan²a+1)
=(1/2+2)/(1/4+1)
=5/2×4/5
=2

If α∈ (π / 2,3 / 2 π) and Tan (α - 7 π) = - 3 / 4, then the value of sin α + cos α is

tan（α-7π）=tana=-3/4=sina/cosa;
Because: Sina ^ 2 + cos ^ 2 = 1, α∈ (π / 2,3 / 2 π) cosa

If Tan (α + 8 π / 7) = a, then: [sin (α + 15 π / 7) + 3cos (α - 13 π / 7)] / [sin (- α + 2 π / 7) - cos (α + 22 π / 7)] = should be discussed in detail Cheng

If x = α + 8 π / 7, then TaNx = a
∴15π/7+α=π+（α+8π/7）=π+x
α-13π/7=(α+8π/7)-3π=x-3π
20π/7-α=4π-(α+8π/7)=4π-x
α+22π/7=(α+8π/7)+2π=x+2π
Thus, on the left side of the original equation:
Left = [sin (π + x) + 3cos (x-3 π)] / [sin (4 π - x) - cos (x + 2 π)]
=(-sinx-3cosx)/(-sinx-cosx)
=(sinx+3cosx)/(sinx+cosx)
=[(sinx+3cosx)/cosx]/[(sinx+cosx)/cosx]
=(tanx+3)/(tanx+1)
=(a+3)/(a+1)

Given that sin α - 3cos α = 0, the value of sin? α + sin α cos α + 2 is

sin²α+sinαcosα+2
=(sin²α+sinαcosα)/(sin²α+cos²α)+2
Substituting sin α - 3cos α = 0, that is sin α = 3cos α
=(9cos²α+3cos²α)/(9cos²α+cos²α)+2
=12/10+2
=6/5+2
=16/5

Given 3cos α - sin α = 1, find the value of sin α cos α

Under the square of both sides, 9cos? α + sin? α - 6sin α cos α = 1 = sin? α + cos? α, i.e. 8cos? α - 6sin α cos α = 0
Elimination of cos α results in 8cos α - 6sin α = 0, so sin α = 4 / 3cos α
The original formula 3cos α - sin α = 3cos α - 4 / 3cos α = 5 / 3cos α = 1
That is cos α = 3 / 5, so sin α = 4 / 3cos α = 4 / 5, so sin α cos α = 4 / 5 * 3 / 5 = 12 / 25

If θ is the second quadrant angle, then θ / 3 cannot be the angle of the second quadrant

Theta / 3 can't be the angle of the second quadrant

Given that the point a (- 3 + A, 2A + 9) is on the bisector of the second quadrant, what is the value of a

Because point a (- 3 + A, 2A + 9) is on the bisector of the second quadrant
So - (- 3 + a) = 2A + 9
a=-6

Given cosa = - 1 / 3, a is the second quadrant angle, sin (a + b) = 1, find the value of COS (2a + b)

sin（a+b）=1
sin²(a+b)+cos²(a+b)=1
So cos (a + b) = 0
cosa=-1/3
sin²a+cos²a=1
A second quadrant
So Sina > 0
sina=2√2/3
The original formula = cos [a + (a + b)]
=cosacos(a+b)-sinasin(a+b)
=-2√2/3

Let α be the angle of the second quadrant, sin α = 3 5, find sin (37 π) 6 − 2 α)

Because sin (37 π)
6−2α)＝sin(π
6−2α)，
sinα＝3
5⇒cosα＝−4
5 (α is Ⅱ)
sin2α＝−24
25cos2α＝1−2sin2α＝7
25 ------ (6 points)
So sin (π)
6−2α)＝7+24
Three
50 -------- (13 points)