# Some basic trigonometry questions

These are all practise questions from my book, i figured out most of them but am stuck on a few. Hoping that someone else can help me

If $\cos(A) + \sin(B) = m$ and $\sin(A) + \cos(B) = n$, prove that $2\sin(a+b) = m^2 + n^2 -2$

I tried to solve it through R.H.S, taking the values of m and n and squaring them.

I got close but i only got $\sin(a+b)$.

I am hoping someone knows how this went wrong, help is highly appreciated!

• Hint: compute $m^2+n^2$ and use $\cos^2\,x+\sin^2\,x=1$. – Raymond Manzoni Aug 7 '13 at 14:22
• I already did that. It finally came to 2 + sin(a+b) -2. Sadly thats sin(a+b) and i need 2sin(a+b) – Aayush Agrawal Aug 7 '13 at 14:23
• Remember that you compute $m^2$ so that the cross-product will be $\;2\cos\,A\sin\,B$. – Raymond Manzoni Aug 7 '13 at 14:24
• oh my god i just realized where i screwed up, thanks alot Raymond! Instead of taking (x+y)^2 as x^2 + 2xy + y^2 i took it as x^2 + xy + y^2 by mistake – Aayush Agrawal Aug 7 '13 at 14:26
• Glad it helped, – Raymond Manzoni Aug 7 '13 at 14:27

I suspect you simply made an arithmetic error. Note that $$m^2+n^2-2 = (\cos^2A+2\cos A\sin B+\sin^2A)+(\sin^2A+2\sin A\cos B+\cos^2B)-2.$$ Can you get the rest of the way using the identities $$\sin(A+B)=\sin A\cos B+\cos A\sin B,$$ and $$\sin^2\theta+\cos^2\theta=1?$$