1. ## trigo proving

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

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

2. Originally Posted by thereddevils
(2) (sin3A+sinA)sinA+(cos3A-cosA)cosA=0
$\displaystyle (\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A$
\displaystyle \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}
\displaystyle \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:

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

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

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

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

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

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

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# SINA SIN3A SIN5A SIN7A = TAN4A

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