Let f (θ) = [sin2 (6 π + θ) + cos θ - 2cos3 (3 π + θ) - 3] / 2 + 2cos2 (θ - 4 π) - cos (- θ), find f (π / 3)

Let f (θ) = [sin2 (6 π + θ) + cos θ - 2cos3 (3 π + θ) - 3] / 2 + 2cos2 (θ - 4 π) - cos (- θ), find f (π / 3)

Are you sure the title is correct
I simplify it to f (θ) = (2cos2 θ + cos θ + 2) (COS θ - 1) / (2 + 2cos2 θ - cos θ), and then I don't need to take it in and calculate it is - 3 / 4

F (&) = 2cos3 & + sin2 (2 π - &) + cos (2 π - &) --- 3 / 2 + 2cos2 (π + &) + cos (- -), find the value of F (π / 3), please 3Q This is a higher mathematical problem, is the induction formula of sine and cosine

The formula = (2cos3 + sin2 + cos + cos + 3) / (2 + 2cos2 + cos + COS) = (2cos3 + 1-cos2 + cos + 3) / (2 + 2cos2 + cos + COS) = (2cos3 + 2 + cos + Cos2) / (2 + 2cos2 + cos + COS) = [2 (COS + 1) (Cos2 + cos + 1) - cos (COS + 1)] / (2 + 2cos2 + cos + COS) / (2 + 2cos2 + COS) / (2 + 2cos2 + cos + COS) = (COS & - 1) (2cos2 + 2cos 2 + 2cos + cos + cos + 1) (2cos2 + 2cos2 + 2cos2 + 2cos 2 + 2cos2 + 2cos2 + 2cos2 + 2cos2 + 2cos 2 + 2cos 2 + by substituting f = (π / 3) into cos & - 1, cos (π / 3) - 1 = 0.5

It is known that Tan α = 1 2, then cos α - sin α 2cosα-sinα= ___ .

∵tanα=1
2，
The original formula = 1-tan α
2-tanα=1-1
Two
2-1
2=1
3．
So the answer is: 1
Three

If sin α + cos α = 7 13, 0 < α < π, then Tan α= Why are sin α and cos α two of the equation x ^ 2 - (7 / 13) X-60 / 169 = 0

Because according to Veda's theorem
Equation x ^ 2 - (7 / 13) X-60 / 169 = 0
x1+x2=-(-7/13)/1=7/13
So sin α and cos α are two of the equations x ^ 2 - (7 / 13) X-60 / 169 = 0

Given sin (π + α) = - 1 / 2, find cos (2 π - α), Tan (α - 7 π)

sin(π+α)=-1/2
It is concluded that a is the first and second quadrant angle
sina=1/2
cosa=±√3/2
tana=±√3/3
cos(2π-a)=cos(-a)=cosa=±√3/2
tan（α-7π）=tana=±√3/3

Given that α + β = 90 ° and sin α + cos β = root 2, find the value of sin? α + 3cos? β

∵ α + β = 90 ° is known
∴β=90°-α ==>cosβ=cos(90°-α)=sinα
∵sinα+cosβ=√2
==>sinα+sinα=√2
==>sinα=√2/2
∴α=β=45°
So sin? α + 3cos? β = (√ 2 / 2) 2 + 3 (√ 2 / 2) 2
=4(√2/2)²
=2.

Tan α is known Tan α − 1 = - 1, find sin α − 3cos α The value of sin α + cos α

∵tanα
tanα−1=-1，
∴tanα=1
2，
∴sinα−3cosα
sinα+cosα=tanα−3
tanα+1=−5
3．

The point m [2, - 4] cannot be in the quadrant

0

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third quadrant

If P (a + 3,15-2a) is on the bisector of the first and third quadrants, then the value of a is What is the bisector of the first and third quadrants,

0