It is known that α is an obtuse angle and Tan (α + π) 4)=-1 7 （Ⅰ）tanα；（Ⅱ）cos2α+1 2cos(α-π 4)-sin2α．

(I) from known: Tan (α + π)
4)=tanα+1
1-tanα=-1
7 (2 points)
Tan α = - 4
3 (5 points)
（Ⅱ）cos2α+1
2cos(α-π
4)-sin2α=2cos2α
sinα+cosα-sin2α=2cos2α
Sin α + cos α - 2Sin α cos α (8 points)
∵α∈(π
2, π) and Tan α = - 4
Three
∴sinα=4
5，cosα=-3
5 (10 points)
∴2cos2α
sinα+cosα-2sinαcosα=2×9
Twenty-five
Four
5-3
5-2×4
5×(-3
5)=18
29 (12 points)

Given that α is an obtuse angle Tan (α + π / 4) = - 1 / 7, find (2) the value of (COS 2 α + 1) / (√ 2 cos (α - π / 4) - sin2 α)

According to the meaning of the title
tana<0
tan(a+π/4)=-1/7
(tana+1)/(1-tana)=-1/7
7tana+7=tana-1
6tana=-8
tana=-4/3
sina>0
So Sina = 4 / 5
cosa<0
cosa=-3/5
(2)cos2a=2cos²a-1=18/25-1=-7/25
√2cos(a-pai/4)=√2(cosa*√2/2+sina*√2/2)=-3/5+4/5=1/5
sin2a=2sinacosa=2*(4/5)*(-3/5)=-24/25
The original formula = (- 7 / 25 + 1) / (1 / 5 + 24 / 25) = (- 7 + 25) / (5 + 24) = 18 / 29

When α∈ (0, π) is known, and sin (π / 3 + α) - sin (π / 3 - α) = 3 / 5, the value of sphere sin2 α - Cos2 α is known

Let sin (α) = 3 / 5, because α ∈ (0, π), cos α = 4 / 5 or - 4 / 5, then the original formula = 31 / 25 or - 3 / 5

Known π / 2

∵π/2

Trigonometric function: F (x) = - √ 3sinx · cosx + cos? X

f(x)=-√3sinx·cosx+cos²x
=-√3/2sin2x+(1+cos2x)/2
=-√3/2sin2x+1/2cos2x+1/2
=sin2xcos(5π/6)+cos2xsin5π/6 +1/2
=sin(2x+5π/6）+1/2
therefore
Max = 1 + 1 / 2 = 3 / 2
Minimum = - 1 + 1 / 2 = - 1 / 2

On trigonometric function: given that (1-cos α) / (1-cos β) = 1 / 9, α + β = 120 °, find α =? β =?

1-cos α = 2 (sin (α / 2)) ^ 21 cos β = 2 (sin (β / 2)) ^ 2, so sin (α / 2) / sin (β / 2) = 1 / 3 (because α / 2, β / 2 are less than 60 degrees, so the sine values are greater than 0) α / 2 = 60 degrees - β / 2 is replaced by sin (α / 2) / Si

Simplified evaluation of trigonometric functions with the same angle Given (TaNx) / (tanx-1) = - 1, calculate the following values: （1）(sinx-3cosx)/(sinx+cosx) （2）(sin^2)x+sinxcosx+2

(tanx)/(tanx-1)=-1
tanx=1-tanx
tanx=1/2
sinx/cosx=tanx=1/2
cosx=2sin
So (sinx-3cosx) / (SiNx + cosx)
=(sinx-6sinx)/(sinx+2sinx)
=-5sinx/3sinx
=-5/3
cosx=2sinx
(sinx)^2+(cosx)^2=1
So (SiNx) ^ 2 = 1 / 5
(cosx)^2=4/5
sinx/cosx=1/2>0
So sinxcosx > 0
Sinxcosx = root (SiNx) ^ 2 * (cosx) ^ 2 = 2 / 5
So (SiNx) ^ 2 + sinxcosx + 2 = 1 / 5 + 2 / 5 + 2 = 13 / 5

Senior one mathematics trigonometric function simplification and evaluation help to think for a long time 3 / 2 * sin ^ 2 20 degrees + 32 * sin ^ 2 20 degrees - 1 / 1 + cos 40 degrees The question on the test paper should not ask you to take the value of sin20 to calculate. I think it should be about 3 / (2 * sin ^ 2 20 degrees) + (32 * sin ^ 2 20 degrees) - 1 / (1 + cos40)

33 (1 / 2) sin ~ 2 20 degrees - 1 / (2-2 sin ~ 2 20 degrees)
Just plug in the value of sin20
Are you sure the title is OK? Why are there such weird numbers as 32 and 3 / 2? I'll think about it again

Proof of simplification and evaluation of trigonometric function 2 Let the fourth angle (sin2) be (sin2) SA / sin2=____________

One
sin(2a+B)-cosB=sin[(a+B)+a]-cos[(a+B)-a]
=sin(a+B)cosa+cos(a+B)sina-cosB
So:
sin(a+B)cosa-1/2[sin(2a+B)-cosB]
=sin(a+B)cosa-1/2sin(a+B)cosa-1/2cos(a+B)sina+1/2cosB
=1/2[sin(a+B)cosa-cos(a+B)sina]+1/2cosB
=1/2(sin(a+B-a)+cosB)
=1/2(sinB+cosB)
=1/2
SINB + CoSb = 1
Squared off, you get sinbcosb = 0
So it can only be CoSb = 0
That is, B = 90 degrees
Two
sin3a=sin(2a+a)=sin2acosa+cos2asina
=2(cosa)^2sina+[1-2(sina)^2]sina
=2sina-2(sina)^3+sina-2(sina)^3
=3sina-4(sina)^3
therefore
(sin3a)/sina=13/5=3-4(sina)^2
The results are as follows:
(sina)^2=1/10
If a is the angle of the fourth quadrant, then Sina = - 1 / Radix 10
tana=-1/3
Then there are:
tan2a=2tana/[1-(tana)^2]
=(-2/3)/[1-(1/9)]
=(-2/3)*(9/8)
=-3/4

Proof of simplification and evaluation of trigonometric function 3 Given that a and B are acute angles, and tanatanb = Tana + tanb + 1, then cos (a + b)=__________

tanAtanB=tanA+tanB+1
tanAtanB-1=tanA+tanB
tan(A+B)=(tanA+tanB)/(1-tanAtanB)=-1
A + B = 135 degrees cos (a + b) = under radical (1 / 2)

It is known that α is an obtuse angle and Tan (α + π) 4)=-1 7 （Ⅰ）tanα； （Ⅱ）cos2α+1 2cos(α-π 4)-sin2α．