Let cos α = - Five 5，tanβ＝1 3，π＜α＜3π 2，0＜β＜π 2. Calculate the value of α - β

Let cos α = - Five 5，tanβ＝1 3，π＜α＜3π 2，0＜β＜π 2. Calculate the value of α - β

∵cosα=-
Five
5，π＜α＜3π
2，
∴sinα=-
1−cos2α=-2
Five
5，
ν Tan α = 2 and Tan β = 1
3，
∴tan（α-β）=tanα−tanβ
1+tanαtanβ=2−1
Three
1+2
3=1，
∵π＜α＜3π
2，0＜β＜π
2，
∴π
2＜α−β＜3π
2，
∴α−β＝5π
4．

If cos α + 2Sin α = - radical 5, then Tan α = a half, B2 C minus half, D negative two

cosα+2sinα=-√5
(cosα+2sinα)²=5
cos²α+4cosαsinα+4sin²α=5
(cos²α+4cosαsinα+4sin²α)/(cos²α+sin²α)=5
The left numerator and denominator of the equation can be obtained by dividing the sum of COS 2 α
(4tan²α+4tanα+1)/(tan²α+1)=5
4tan²α+4tanα+1=5(tan²α+1)
(tanα-2)²=0
tanα=2
Hope to adopt, if you do not understand, please ask

Given that Tan α = 1 / 7, sin β = 10 / 10, find: (1) the value of sin α, (2) the value of COS (α + 2 β), Thank you very much for the answers. The ideas are OK

If you draw a right triangle, one right angle side is 7, and the other right angle side is 1, the oblique side is the root sign 50, so you can find sin α. As for cos (α + 2 β), you should first split Cos2 β, and use the angle doubling formula

Sin (α + π / 3) + sin α = the root of minus 2, 3 α ∈ (- π / 2,0), find cos α How is it that three times sin α plus half times cos α equals five quarters of minus?

sin(α+π/3) + sinα = -√3/2
sinαcosπ/3 + cosαsinπ/3 + sinα = -√3/2
1/2sinα + sinα + √3/2cosα = -√3/2
Because sin α ^ 2 + cos α ^ 2 = 1
So according to these two formulas, we can calculate cos α = - 1 or 1 / 2
Because α∈ (- π / 2,0), cos α = 1 / 2

It is known that cos (π - 2 α) / sin (α - (π / 4)) = - the root two of negative half, find sin α + cos α=

cos(π-2α)/sin(α-(π/4)) = -√2/22cos2α =√2sin(α-(π/4))=√2(sinαcosα(π/4)-cosαsin(π/4))= sinα-cosα2((cosα)^2-(sinα)^2 = sinα-cosα2(cosα+sinα)(cosα-sinα)= sinα-cosαsinα+cosα = ...

Let X be the third quadrant angle 1−cos2x=（　　） A. 2sinx B. − 2sinx C. 2cosx D. − 2cosx

1−cos2x=
1−(1−2sin2x )=
2|sinx|，
Because x is the third quadrant angle, SiNx < 0,
So the above formula = −
2sinx．
Therefore, B

In the triangle ABC, if sin (2 pai-a) = - heel 2Sin (pai-b) and 3cosa = - heel 2cos (pai-b), find the three inner angles of the triangle

The first formula is converted into sin a = √ 2 sin B,
The second formula is reduced to: √ 3 cos a = √ 2 cos B,
Multiply the two formulas to get:
√(3/2) sin(2A)=sin(2B),
Because of a + B

In the triangle ABC, if sin (2 π - a) = - √ 2Sin (π - b), √ 3cos (2 π - a) = - √ 2cos (π + b), find the three angles of the triangle

sin（2π-A）=-√2sin（π-B）
-sinA=-√2sinB
sinA=√2sinB (1)
√3cos（2π-A）=-√2cos（π+B）
√ 3cosa = √ 2cosb (A and B must be acute angles to be equal)
cosA=(√6/3)cosB (2)
(1) ^ 2 + (2) ^ 2
2(sinB)^2+2/3(cosB)^2=1
2-2(cosB)^2+2/3(cosB)^2=1
4/3(cosB)^2=1
cosB=√3/2 B=π/6
cosA=√2/2 A=π/4
C=π-A-B=7π/12

In triangle ABC, if sin (2 π + a) = √ 2Sin (π - 2b), √ 3cosa = - √ 2cos (π - b), calculate the degree of each inner angle of triangle ABC

Sin (2 π - a) = √ 2Sin (π + b), namely: Sina = √ 2sinb ------ - (1) √ 3cosa = - √ 2cos (π - b), namely: cosa = √ (2 / 3) CoSb ------ (2) (1) ^ 2 + (2) ^ 2, the result is: 1 = 2 (SINB) ^ 2 + 2 / 3 (CoSb) ^ 2, the deformation is: 3 = 6 (SINB) ^ 2 + 2 (CoSb) ^ 2 = 2 + 4 (SINB) ^ 2

Simplify sin (x + 60 °) + 2Sin (X-60 °)- The results of 3cos (120 ° - x) are as follows______ .

The original formula = sin (x + 60 °)-
3cos[180°-（x+60°）]+2sin（x-60°）
=sin（x+60°）+
3cos（x+60°）+2sin（x-60°）
=2sin（x+60°+60°）+2sin（x-60°）
=2sin（x-60°+180°）+2sin（x-60°）
=-2sin（x-60°）+2sin（x-60°）
=0．
So the answer is 0