# math is a PROBLEM!

• Apr 21st 2008, 01:09 PM
Jono Mathematician
math is a PROBLEM!
sinA

=

cosA-coA.cos2A-(1/2)sin2A
sin2A-cosA

ALSO

8sinA.cosA.cos2A.cos4A = sin8A

how do you do these? and can u explain the 2nd one as well ... thanks
• Apr 21st 2008, 02:27 PM
wingless
Use $\sin 2x = 2\sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$.
• Apr 21st 2008, 03:19 PM
Jono Mathematician
yeah i know that silly, but how i do it. haha
• Apr 21st 2008, 08:20 PM
Soroban
Hello, Jono Mathematician!

Wingless suggested a few identities:

. . $\sin2\theta \:=\:2\sin\theta\cos\theta\;\;{\color{blue}[1]}$
. . $\cos2\theta \:=\:1-2\sin^2\!\theta\;\;{\color{blue}[2]}$

We must find ways to apply them . . .

Quote:

$\frac{\cos A - \cos A\cos2A - \frac{1}{2}\sin2A}{\sin2A-\cos A} \;=\;\sin A$
The left side is: . $\frac{\cos A - \cos A\overbrace{\cos2A}^{[2]} - \frac{1}{2}\overbrace{\sin2A}^{[1]}}{\underbrace{\sin2A}_{[1]}-\cos A} \;= \;\frac{\cos A - \cos A\overbrace{(1 - 2\sin^2\!A)} - \frac{1}{2}\overbrace{(2\sin A\cos A)}}{\underbrace{2\sin A\cos A} - \cos A}$

. . $= \;\frac{\cos A - \cos A + 2\sin^2\!A\cos A - \sin A\cos A}{2\sin A\cos A - \cos A} \;=\;\frac{2\sin^2\!A\cos A - \sin A\cos A}{2\sin A\cos A - cos A}$

. . $= \;\frac{\sin A\!\cdot\!\cos A\!\cdot\!(2\sin A - 1)}{\cos A\!\cdot\!(2\sin -1)} \;=\; \sin A$

Quote:

$8\sin A \cos A \cos2A\cos4A \:= \:\sin8A$
$\text{The left side is: }\;4\underbrace{(2\sin A\cos A)}\cos2A\cos4A$

. . . . . . . . . $= \qquad4\sin2A\cos2A\cos4A$

. . . . . . . . . $= \;2\underbrace{(2\sin2A\cos2A)}\cos4A$

. . . . . . . . . $=\qquad\underbrace{2\sin4A\cos4A}$

. . . . . . . . . $= \;\qquad\quad\sin8A$