Evaluation: cos (π / 13) + cos (3 π / 13) + cos (9 π / 13) Of course, the result I want is not the approximate value, but the exact value! At the same time, there should be a brief problem-solving process.

Evaluation: cos (π / 13) + cos (3 π / 13) + cos (9 π / 13) Of course, the result I want is not the approximate value, but the exact value! At the same time, there should be a brief problem-solving process.

(1 + radical 13) / 4 it seems that the upper and lower letters of the summation sign are not displayed correctly. Anyway, the upper part is n and the lower part is k = 1. For convenience, remember cos (π / 13) + cos (3 π / 13) + cos (9 π / 13) is (1,3,9) 13, let (1,3,9) 13 = x, (5,7,11) 13 = y (i.e. cos (5 π / 13) + cos (7 π / 13) + cos (11 π / 13)

If Cos2 α + 2Sin β = 3, then α=

∵cos2α≤1,2sinβ≤2
∴cos2α+2sinβ≤3
But Cos2 α + 2Sin β = 3
∴cos2α=sinβ=1
∴2α=2kπ
ν α = k π (k is an integer)
Do not know, welcome to ask

cos2α-cos2β=2sin(α+β)sin(β-α) I can't understand the answer. It seems that there is no formula

Let a = α + β, B = β - α,
Then 2 α = A-B, 2 β = a + B
So Cos2 α - Cos2 β = cos (a-b) - cos (a + b)
=cosAcosB+sinAsinB-( cosAcosB-sinAsinB )
=2sinAsinB
=2sin(α+β)sin(β-α)

Given α∈ (0, π / 2), and 2Sin α - sin α cos α - 3cos α = 0. Find the value of [sin (α + π / 4)] / (sin2 α + Cos2 α + 1)

2 sin α - 3 cos α = 0 get (2 sin α - 3 cos α) (sin α + cos α) = 0 sin2 α + cos 2 α + 1 = 2cos α (sin α + cos α) get sin α + cos α = 0, so 2Sin α - 3cos α = 0, so tan α = 3 / 2 get cos α = 2 / (√ 13) (sin (α + π / 4)) / (sin2 α + cos 2 α + 1) = ((√ 2) / 2) (2) (sin α + cos α) / (2cos α + cos α) / (2cos α + cos α) / (2cos α (sin α (SIN) α (sin α + cos 2 α + 1) = (√ 2) / 2) (2) (sin α + cos α) / (2cos α (sin α (sin α (sin α) α (sin α (sin α) α + cos α)) = (√ 26) / 8

(2sin2 α / 1 + Cos2 α) * (COS α) ^ 2 / Cos2 α = what

(2sin2α/1+cos2α)*(cosα)^2/cos2α
=(2sin2α)/(2cos²α) *(cos²α)/(cos2α)
=sin2α/cos2α
=tan2α

Simplify 2sin2 α 1+cos2α•cos2α cos2α=______ .

Because Cos2 α = 2cos2 α - 1
Therefore, the original formula = 2sin2 α
1+(2cos2 α−1)•cos2α
cos2α＝2sin2α
2cos2α•cos2α
cos2α＝sin2α
cos2α＝tan2α
The answer is: α

Given π < θ < 3 π / 2, cos θ = - 4 / 5, find cos θ / 2

Cos θ = - 4 / 5 = 2cos ^ 2 (θ / 2) - 1, cos (θ / 2) = ± √ 10 / 10 π / 2 ＜ θ / 2 ＜ 3 π / 4, cos (θ / 2) ＜ 0, so cos (θ / 2) = - √ 10 / 10

If cos α = - 3 / 5, α belongs to (π / 2,2 π), then cos (α - π / 4)=

Cos α = - 3 / 5, α belongs to (π / 2,2 π)?
It should be that α belongs to (π / 2, π), otherwise the range is meaningless
sinα>0
Sin 2 α = 1-cos 2 α = 1 - (- 3 / 5) 2 = 16 / 25
∴ sinα=4/5
∴ cos(α-π/4)
=cosα*cos(π/4)+sinαsin(π/4)
=(-3/5)*(√2/2)+(4/5)*(√2/2)
=√2/10

Given cos α = 4 / 5, α∈ (3 π, 2 π), find Tan α

cos^2α+sin^2α=1
Divide both sides by cos ^ 2 α at the same time
The original formula becomes 1 + Tan ^ 2 α = 1 / cos ^ 2 α = 25 / 16
So tan ^ 2 α = 9 / 16
α∈ (2 π, 3 π), cos α = 4 / 5 is greater than 0
So alpha is in the first quadrant
So tan α = 3 / 4

It is known that cos θ = - 3 / 5, and π

From cosa we can calculate Sina
You can calculate Tana
Then we expand Tan (θ = π / 4) =
I'm bringing it in
That's it
I hope it can help you