establishing identities?

**pongpong** · February 22nd 2009, 11:18 PM

how to prove?

(cot^2 x-1) / (2 cot x) = cot 2x

cos 3x = 4 cos^3 x-3 cosx

sin 2t+sin t/1+cos t+sos 2t = tan t

THANK YOU!

**mathaddict** · February 22nd 2009, 11:46 PM

$\frac{cot^2x-1}{2cotx}$

$=\frac{\frac{1-tan^2x}{tan^2x}}{\frac{2}{tanx}}$

$=\frac{1-tan^2x}{2tanx}$

$=cot2x$

Knowing the fact that cot x = 1/tan x .

**mathaddict** · February 22nd 2009, 11:56 PM

Originally Posted by pongpong

how to prove?

cos 3x = 4 cos^3 x-3 cosx

cos(2x+x)=cos2xcosx-sin2xsinx

=(2cos^2x-1)cosx-2sinxcosx(sinx)

=2cos^3x-cosx-2sin^2xcosx

=2cos^3x-cosx-2(1-cos^2x)(cosx)

=2cos^3x-cosx-2cosx+2cos^3x

=4cos^3x-3cosx

**Soroban** · February 23rd 2009, 05:55 AM

Hello, pongpong!

The third one is impossible to read.
Well, not "impossible" . . . I figured it out!

$\frac{\sin2t+\sin t}{1+\cos t+\cos 2t} \:=\: \tan t$

We're expected to know these Double-Angle Identities:

. . . . . $\begin{array}{ccc}\sin2\theta &=& 2\sin\theta\cos\theta \\ \cos2\theta &=& 2\cos^2\!\theta - 1\end{array}$

The left side becomes: . $\frac{2\sin t\cos t + \sin t}{1 + \cos 2 + 2\cos^2\!t - 1} \;=\;\frac{2\sin t\cos t + \sin t}{2\cos^2\!t + \cos t}$

Factor and reduce: . $\frac{\sin t(2\cos t + 1)}{\cos t(2\cos t - 1)} \;=\;\frac{\sin t}{\cos t} \;=\;\tan t$

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