Hello rolisp
Welcome to Math Help Forum! Originally Posted by
rolisp cosA+cos2A+cos6A+cos7A= 4cosA/2 cos5A/2 cos4A
(sinA-sinB)(sinA+sinB)= sin(A-B) sin(A+B)
(sin2A+sin5A-sin3A)/(cosA+1-2sin^2 2A) = 2sinA
I have no idea how to prove these identities. Could anyone help me saying any idea how to prove? Thanks
In each of these questions you need to use the Sum-to-Product formulas (and in #2 and #3, the double-angle formulae) together with basic factorising and simplifying techniques.
I'll start you off in each case, and leave you to try to complete it.
1) $\displaystyle \cos A +\cos 2A+\cos 6A+\cos 7A$$\displaystyle =2\cos \Big(\frac{3A}{2}\Big)\cos \Big(\frac{A}{2}\Big)+2\cos \Big(\frac{13A}{2}\Big)\cos \Big(\frac{A}{2}\Big)$
Now factorise by taking out a common factor $\displaystyle 2\cos \Big(\frac{A}{2}\Big)$. Then use the same sum-to-product formula again, and you're there.
2) $\displaystyle (\sin A - \sin B)(\sin A + \sin B)$$\displaystyle = 2\cos \Big(\frac{A+B}{2}\Big)\sin\Big(\frac{A-B}{2}\Big)\times 2\sin\Big(\frac{A+B}{2}\Big)\cos\Big(\frac{A-B}{2}\Big)$
Now re-arrange, and use the identity $\displaystyle \sin2\theta=2\sin\theta\cos\theta $ (twice), to write the expression in terms of $\displaystyle \sin(A+B)$ and $\displaystyle \sin(A-B)$.
3) $\displaystyle \frac{\sin2A+\sin5A-\sin3A}{\cos A +1 -2\sin^22A}$
$\displaystyle =\frac{2\sin A \cos A +2\cos4A\sin A}{\cos A + \cos 4A}$
Factorise; cancel and you're there.
Let us know how you get on.
Grandad