1. trigo proving

(1) $
\frac{sinA+sin3A+sin5A+sin7A}{cosA+cos3A+cos5A+cos 7A}=tan4A
$

(2) (sin3A+sinA)sinA+(cos3A-cosA)cosA=0

2. Originally Posted by thereddevils
(2) (sin3A+sinA)sinA+(cos3A-cosA)cosA=0
$(\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A$
\begin{aligned}
&= \sin A \sin 3A + \sin^2 A + \cos A \cos 3A - \cos^2 A \\
&= \sin A (3 \sin A - 4 \sin^3 A) + \sin^2 A + \cos A (4\cos^3 A - 3\cos A) - \cos^2 A \\
&= 3\sin^2 A - 4\sin^4 A + \sin^2 A + 4\cos^4 A - 3\cos^2 A - \cos^2 A \\
&= -4\sin^4 A + 4\sin^2 A + 4\cos^4 A - 4\cos^2 A \\
&= 4(\cos^4 A - \sin^4 A) + 4\sin^2 A - 4\cos^2 A
\end{aligned}

\begin{aligned}
&= 4(\cos^2 A + \sin^2 A)(\cos^2 A - \sin^2 A) + 4\sin^2 A - 4\cos^2 A \\
&= 4(\cos^2 A - \sin^2 A) + 4\sin^2 A - 4\cos^2 A \\
&= 4\cos^2 A - 4\sin^2 A + 4\sin^2 A - 4\cos^2 A \\
&= 0
\end{aligned}

01

3. Helo, thereddevils!

We need two Sum-to-Product identities:

. . $\begin{array}{ccc}\sin A + \sin B &=& 2\sin(\frac{A+B}{2}) \cos(\frac{A-B}{2}) \\ \\[-3mm] \cos A + \cos B &=& 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) \end{array}$

$1)\quad \frac{\sin A+\sin3A+\sin5A+\sin7A}{\cos A+\cos3A+\cos5A+\cos7A}\;=\;\tan4A$
Numerator: . $(\sin A + \sin 7A) + (\sin3A + \sin5A) \;=\;2\sin4A\cos3A + 2\sin4A\cos A$

. . Factor: . $2\sin4A(\cos3A + \cos A)$

Denominator: . $(\cos A + \cos7A) + (\cos3A + \cos5A) \;=\;2\cos4A\cos3A + 2\cos4A\cos A$

. . Factor: . $2\cos4A(\cos3A + \cos A)$

The fraction becomes: . $\frac{2\sin4A\,(\cos3A + \cos A)}{2\cos4A\,(\cos3A + \cos A)} \;=\;\frac{\sin4A}{\cos4A} \;=\;
\tan4A$

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(sin3A 2sin5A sin7A)/(sin3A.sin4A - Cos2A.cosA)

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