Question

Prove that:
cos(A+B)+sin(A-B)=2sin(45+A)cos(45+B)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@Neo92 ok i will solve

Answer 5

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)

Answer 6

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)

Answer 7

* http://www.goiit.com/posts/list/trignometry-cos-a-b-sin-a-b-2sin-45-a-cos-45-b-913366.htm;jsessionid=FFECB675EAB95823CBAB29FA987B3A79.node2#943927

Answer 8

\[\cos(A+B)+\sin(A-B)=\cos A\cos B-\sin A\sin B+\sin A \cos B-\sin B\cos A=\]
\[=\cos A(\cos B-\sin B)+\sin A (\cos B-\sin B)=(\cos A+\sin A)(\cos B-\sin B)\]
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:
\[(\cos A+\sin A)(\cos B-\sin B)=\sqrt2\sin(45^\circ +A)\cdot \sqrt2\cos(45^\circ +B)=2\sin(45^\circ +A)\cdot\]\[\cdot\cos(45^\circ +B)\]