It is known that α is the angle of the third quadrant, and f (α) = sin (π − α) cos (2 π − α) Tan (3 π) 2−α) cos(−π−α)． (1) Simplify f (α) (2) If cos (α - 3 π) 2）=1 5. Find the value of F (α); (3) If α = - 1860 °, find the value of F (α)

It is known that α is the angle of the third quadrant, and f (α) = sin (π − α) cos (2 π − α) Tan (3 π) 2−α) cos(−π−α)． (1) Simplify f (α) (2) If cos (α - 3 π) 2）=1 5. Find the value of F (α); (3) If α = - 1860 °, find the value of F (α)

(1) The results show that f (α) = sin α cos α sin (3 π 2 − α) − cos α cos (3 π 2 − α) = sin α cos α (− cos α) − cos α (− sin α) = - cos α; (2) ∵ cos (α - 3 π 2) = 15, ? cos (α - 3 π 2) = 15, ? cos (α - 3 π 2) = 15

Given that α is the third quadrant angle, f (α) = [sin (π - a) cos (2 π - a) Tan (1.5 π - a) Tan (- A - π)] / sin (- π - a), simplify f (α)

sin(π-a)=sina
cos(2π-a)=cosa
tan(1.5π-a)=cota
tan(-a-π)=-tana
sin(-π-a)=sina
So f (a) = sinacosacota (- Tana) / sin
=-cosa

Let y = sin (COS α) / cos (sin α), and α be the second quadrant angle, judge its sign

∵ α is the second quadrant angle
∴-1

Given tan2 θ = - 2 radical 2, π〈 2 θ, find (2cos ^ 2 θ / 2-sin θ - 1) / (Radix 2Sin (θ + 9 π / 4))

Tan2, θ = - 2, radical 2, π

If a ∈ (π / 2, π), sin α = radical five / 5, then tan2 α=

A ∈ (π / 2, π), then cosa

If sin α - 2cos α = 2 / 2 root 10, then tan2 α=

If the square of two sides is the same time, we can get:
(sinα)^2-4sinαcosα+4(cosα)^2=5/2
1-2sin2α+3(cosα)^2=1+3/2
3(cosα)^2-2sin2α=3/2
∵cos2α=2(cosα)^2-1
ν 1.5cos2 α = 3 (COS α) ^ 2-1.5, where two edges are multiplied by 1.5 at the same time, the following formula is obtained:
3(cosα)^2=1.5+1.5cos2α
1.5=3/2
Then there are
1.5+1.5cos2α-2sin2α=3/2
1.5cos2α=2sin2α
3cos2α=4sin2α
tan2α=3/4

Given that tan2 α = - 2 radical 2 and Tan α > 1, calculate the value of [2cos square α / 2-sin α - 1] / [root 2Sin (π / 4 + α)]

Analysis: ∵ tan2a = 2tana / [1 - (Tana) ^ 2] = - 2 √ 2,
∴√2(tana)^2-tana-√2=0,
/ / Tana = √ 2, Tana = - √ 2 / 2 (omitted)
[2(cosa/2)^2-sina-1]/[√2sin(π/4+α）]
=(cosa-sina)/(sina+cosa)
=（1-tana)/(1+tana)
=(1-√2)/(1+√2)
=(1-√2)^2/(1-2)
=2√2-3

Given that 0 ° < θ < 90 ° and sin θ - Radix 2cos θ = 0, we can find the value of 2Sin θ + cos θ / 2Sin θ - cos θ

Sin θ - √ 2cos θ = 0, sin θ = √ 2cos θ
So there are:
(2sinθ+cosθ)/(2sinθ-cosθ)
=(2√2cosθ+cosθ)/(2√2cosθ-cosθ)
=(2√2+1)/(2√2-1)
=(9+4√2)/7

How to deduce radical 3 / 2cos (B / 2) = 2Sin (B / 2) cos (B / 2)

A. Are the three angles of B and C in a triangle? The root sign 3 / 2 is transformed into cos (π / 6), 2Sin (B / 2) cos (B / 2) = sin (b), π / 6 = (A-C) / 2, and then s = 2sina * BC = 2sinb * AC = 2sinc * AB is deduced according to the area company
The following conditions are as follows: radical 3 / 2cos (B / 2) = cos (π / 6) * cos (B / 2) = cos ((A-C) / 2) * cos ((π - A-C) / 2) = cos ((A-C) / 2) * sin ((a + C) / 2) = (Sina + sinc) / 2 = sin (b), a + C = 2B. This condition is very important

Given the vector a = (2cos α, 2Sin α) B = (sin β, cos β), find the minimum value of the modulus of vector a + B if vector C = (- 1 / 2, root 3 / 2) and vector a * B And the vector a * b = 3 / 5 β∈ (0, π) is used to find the value of sin β

P is α, q is β: | a + B | 2 = (a + b) · (a + b) = (a + b) · (a + b) = (a + b) | (a + b) = (a + b) | (a + b) = | a | a + B | 2 + 2 + 2A · B = 4 + 1 + 2 (2cosp, 2sinp) · (SINQ, cosq) = 5 + 4 (cosp SINQ + sinpcoq) = 5 + 4sin (P + Q) = 5 + 4sin (P + Q), therefore: | a + B | 2 | 2 ∈ [1,9], that is: the minimum value of a + B | is 1a · B | 1a · B = (2cosp, 2sinp) · (SINQ, cosq) = 2Sin (P