Prove: sin2a + sin2b-sin2a * sin2b + cos2a * cos2b = 1

Prove: sin2a + sin2b-sin2a * sin2b + cos2a * cos2b = 1

If a > b in triangle, how to prove sin2a > sin2b, and cos2a

False proposition! When a = 75 ° and B = 60 °
sin2A=1/2,sin2B=√3/2

sin2a=sin2b
Then a = B or a + B = π / 2, how to prove the conclusion

This should be the inner angle of the triangle
sin2a=sin2b
sinx=sin(π-x)
sin2b=sin(π-2b)=sin2a
So 2A = 2B or 2A + 2B = π
So a = B or a + B = π / 2

Why can sin2a = sin2b introduce 2cos (a, b) sin (a – b) = 0

sin2A=2sinAcosA;sin2B=sinBcosB
Because sin2a = sin2b; so case 1 ` Sina = cosa, SINB = CoSb; or
Case 2 Sina = CoSb, cosa = SINB
But 2cos (a + b) · sin (a-b) = 2 (cosacosb sinasinb) (sinacosb sinbcosa)
In case 1, cosacosb sinasinb = 0;
In case 2, sinacosb sinbcosa = 0
So 2cos (a + b) · sin (a-b) = 0

If Tan (a + b) = 3tana, prove 2sin2b-sin2a = sin (2a + 3b)

It's like 2sin2b-sin2a = sin (2a + 2b)

The proof of sin α = 2Sin (2a + α): Tan (a + α) = - 3tana

That is sin [(a + α) - A] = 2Sin [(a + α) + a]
sin(a+α)cosa-cos(a+α)sina=2sin(a+α)cosa+2cos(a+α)sina
sin(a+α)cosa=-3cos(a+α)sina
Two sides divided by cosacos (a + α)
sin(a+α)/cos(a+α)=-3cos(a+α)sina/cosa
tan(a+α)=-3tana

It is proved that 2Sin (a + b) cos (a-b) = sin2a + sin2b

Proof from right to left
sin2a+sin2b=sin[（a+b)+(a-b）]+sin[(a+b)-(a-b)]
Then (a + b) and (a-b) are regarded as a whole, and the left is obtained

How to prove 2Sin (π + a) cos (π - a) = sin2a

2sin(π+a）cos（π-a）=(-2sina)(-cosa)=sin2a

Why sin2a + sin2b can be reduced to 2Sin (a + b) cos (a-b)

sin2A+sin2B=sin((A+B)+(A-B))+sin((A+B）-（A-B))=sin(A+B)cos(A-B)+cos(A+B)sin(A-B)+sin(A+B)cos(A-B)-cos(A+B)sin(A-B)=2sin(A+B)cos(A-B)

Sin α = 4 / 5, then sin (α + π / 4) + cos (α + π / 4) equals α, which is the angle of the second quadrant
What is the sum of (α is the angle of the second quadrant) sin (α + π / 4) + cos (α + π / 4)

The original formula is sin α cos π / 4 + cos α sin π / 4 + cos α cos π / 4-sin α sin π / 4
Then bring it in