## shubham.bagrecha 3 years ago Prove that: cos(A+B)+sin(A-B)=2sin(45+A)cos(45+B)

1. shubham.bagrecha

$\cos(A+B)+\sin(A-B)=2\sin(45+A)\cos(45+B)$

2. Neo92

first write cos(A+B) as sin[90-(A+B)] then use sinC + sinD = 2sin[(C+D)/2]cos[(C-D)/2] try and tell me what u get

3. mathslover

ok ... for this first of all tell me what is cos(A+B)

4. shubham.bagrecha

@Neo92 ok i will solve

5. mathslover

cos(A+B)=sin(90-A-B) sin(90-A-B) + sin(A-B) = 2sin(45+A)cos(45+B) we know that sin(x) + sin(y)=2sin([x+y]/2)cos([x-y]/2)

6. mathslover

so we get that sin(90-A-B) + sin(A-B)=2sin(45-B)cos(45-A) but we know that sin(45-B)=cos(90-(45-b))=cos(45+B) and cos(45-A)=sin(90-(45-A))=sin(45+A) hence we get that cos(A+B) + sin(A-B) = 2sin(45+A)cos(45+B)

7. mathslover
8. klimenkov

$\cos(A+B)+\sin(A-B)=\cos A\cos B-\sin A\sin B+\sin A \cos B-\sin B\cos A=$ $=\cos A(\cos B-\sin B)+\sin A (\cos B-\sin B)=(\cos A+\sin A)(\cos B-\sin B)$ Now use: $$\cos A+\sin A=\sqrt2\cdot(\frac1{\sqrt2}\cos A+\frac1{\sqrt2}\sin A)=\sqrt2\sin(45^\circ +A)$$ and $$\cos B-\sin B=\sqrt2\cdot(\frac1{\sqrt2}\cos B-\frac1{\sqrt2}\sin B)=\sqrt2\cos(45^\circ +B)$$. Now multiply this: $(\cos A+\sin A)(\cos B-\sin B)=\sqrt2\sin(45^\circ +A)\cdot \sqrt2\cos(45^\circ +B)=2\sin(45^\circ +A)\cdot$$\cdot\cos(45^\circ +B)$

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