The detailed process of 1 + sin θ - cos θ / 1 + sin θ + cos θ is simplified

The detailed process of 1 + sin θ - cos θ / 1 + sin θ + cos θ is simplified

Application formula:
sina=2sina/2*cosa/2
cosa=2(cosa/2)^2-1=1-2(sina/2)^2
（1+sinθ-cosθ0/(1+sinθ+cosθ）
=(1+2sinθ/2*cosθ/2-1+2sin^2θ/2)/(1+2sinθ/2*cosθ/2+2cos^2θ/2-1)
=(2sinθ/2*cosθ/2+2sin^2θ/2)/(2sinθ/2*cosθ/2+2cos^2θ/2)
=2sinθ/2(sinθ/2+cosθ/2)/2cosθ/2(sinθ/2+cosθ/2)
=sinθ/2/cosθ/2
=tanθ/2

[sin (5 π - α) cos (- π - α)] / [cos (α - π) cos (π / 2 + α)] simplification

sina×(-cosa)/-cosa×（-sina）=-1

Simplification: sin (a + 5 π) cos (- π / 2-A) · cos (8 π - a) / sin (A-3 π) · sin (- A-4 π) Simplification: sin (a + 5 π) cos (- π / 2-A) · cos (8 π - a) / sin (A-3 π / 2) · sin (- A-4 π)

sin(a+5π)=-sina,cos(-π/2-a)=cos(π/2+a)=-sina,cos(8π-a)=cosa；
sin(a-3π/2)=cosa,sin(-a-4π)=sina；
The original formula = (- Sina) (- Sina) cosa / (cosasina) = Sina;

Simplification of sin (a-5 π) cos (- π / 2-A) cos (8 π - a) / sin (A-3 π / 2) sin (- A-4 π)

=-sinα*sinα*-cosα/cosα*-sinα
=-sinα

(2 cosa Sina) (Sina + cosa + 3) = 0, then (2cos? 2A + sin2a) / (1 + Tana) =?

(2cosa-sina)(sina+cosa+3)=0
If Sina + cosa + 3 = 0, Sina + cosa = - 3
Because | Sina | 1 | Cosa | 1
sina+cosa≥-2
So this is not true
If 2cosa = Sina
Substituting (COSA) ^ 2 + (Sina) ^ 2 = 1
(cosa)^2=1/5
[2(cosa)^2+sin2a]/(1+tana)
=[2(cosa)^2+2sinacosa]/[(sina+cosa)/cosa]
=[2cosa(sina+cosa)]/[(sina+cosa)/cosa]
=2(cosa)^2
=2/5

Let f (x) = sinx-2cosx get the maximum value when x = θ, then cos θ = 0___ .

f（x）=sinx-2cosx=
5（
Five
5sinx-2
Five
5cosx）=
5sin (x - α) (where cos α=
Five
5，sinα=2
Five
5），
When ∵ x = θ, the function f (x) gets the maximum value,
ν sin (θ - α) = 1, namely sin θ - 2cos θ=
5，
In addition, sin2 θ + Cos2 θ = 1, the simultaneous solution gives cos θ = - 2
Five
5．
The answer is as follows:
Five
Five

F (x) SiNx · sin (π - x) + √ 3sin (π / 2 + x) cos (π / 2 + x) + 2cos (π + x) cos (π - x) to find the minimum positive period of FX

SiNx (x) = SiNx · sin (π - x) + 3 / 3 sin (π / 2 + x) cos (π / 2 + x) + 2cos (π + x) cos (π - X - x) = sin ^ 2x - √ 3cos xsinx + 2cos ^ 2x = cos ^ 2x - √ 3cos xsinx + 1 = (cos2x + 1) / 2 - √ 3 / 2 * sin2x + 1 = 1 / 2cos2x - √ 3 / 2 * sin2x + 3 / 2 * sin2x + 3 / 2 * sin2x + 3 / 2 * sin2x + 3 / 2 * sin2x + 3 / 2 = cos (2x + π / 3) + 3 / 2 = cos (2x + π / 3) + 3 / 2 = cos (2x + π / the minimum positive period of 3 / 2fx is

If the point P (1,2) is known to be on the final edge of α, then 6sin α + cos α / 3sin α - 2cos α= My answer is 13 / 4

Using the definition of tangent function
tanα=y/x=2
(6sinα+cosα)/(3sinα-2cosα)
Denominator divided by cos α at the same time
=(6tanα+1)/(3tanα-2)
=(6*2+1)/(3*2-2)
=13/4
Your answer is right
It seems that your study is very good

Given cos (α - 9 π / 2) = 2cos (5 π - α), find the value of sin (α - π) ^ 2 + 3sin (2 π - α) cos (5 π + α)

cos(α-9π/2）=cos(a-π/2)=cos(π/2-a)=sina2cos(5π-α)=2cos(π-a)=-2cosasina=-2cosa(sina)^2+(cosa)^2=5(sina)^2/4=1 (sina)^2=4/5 (cosa)^2=1/5sin(α-π）^2+3sin(2π-α）cos(5π+α）=(sina)^2+3(-sina)(-...

If Tan α = 2, then 3sin ^ 2 α - sin α cos α - 2cos ^ 2 α =?

3sin^2α-sinαcosα-2cos^2α
=(3sin^2α-sinαcosα-2cos^2α)/1
=(3sin ^ 2 α - sin α cos α - 2cos ^ 2 α) / (sin ^ 2A + cos ^ 2a)
=(3tan^2a-tana-2)/(tan^2a+1)
=12/5