Very Difficult Trig Identity Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

**D4mien** · April 18th 2008, 02:27 PM

Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

Here are the Identities I (read: You) have to work with

cscX = 1/sinX
secX = 1/cosX
cotX = 1/tanX

tanX = sinX/cosX
cotX = cosX/sinX

sin(x+y) = sinXcosY + cosXsinY
sin(x-y) = sinXcosY - cosXsinY

cos(x+y) = cosXcosY - sinXsinY
cos(x-y) = cosXcosY + sinXsinY

tan(x+y) = tanX+tanY/1-tanXtanY
tan(x-y) = tanX-tanY/1+tanXtanY

cos2X = (cos(X))^2 - (sin(x))^2
cos2X = 1-2(sin(x))^2
cos2X = 2(cos(x))^2 - 1

sin2X = 2sinXcosX

tan2X = 2tanX/1- (tan(x))^2

**mrbuttersworth** · April 18th 2008, 05:47 PM

Hi, I tried this one for a while with no luck. But just from plugging it into a calculator, I'm skeptical that it is in fact an identity.

$cot(2x) - cot(3x) = \frac{1}{2sec(x)csc(3x)}$

Are not equal when put into a calculator. Maybe I plugged the numbers in incorrectly, or maybe I misinterpreted your formula, but as far as I know, the above is not an identity. Could I get verification on this?

**Soroban** · April 18th 2008, 06:16 PM

Hello, D4mien!

$\cot(2x) - \cot(3x) \:= \:\frac{1}{2}\sec x\csc 3x$

$\cot(2x) - \cot(3x) \;=\; \frac{\cos(2x)}{\sin(2x)} - \frac{\cos(3x)}{\sin(3x)} \;=\;\frac{\overbrace{\sin(3x)\cos(2x) - \sin(2x)\cos(3x)}^{\text{This is }\sin(3x-2x)}}{\sin(2x)\sin(3x)}$

. . . . . $= \;\frac{\sin(x)}{\sin(2x)\sin(3x)} \;=\;\frac{\sin(x)}{2\sin(x)\cos(x)\cdot\sin(3x)} \;=\;\frac{1}{2\cos(x)\sin(3x)}$

. . . . . $= \;\frac{1}{2}\cdot\frac{1}{\cos(x)}\cdot\frac{1}{\ sin(3x)} \;=\;\frac{1}{2}\sec(x)\csc(3x)$

**mrbuttersworth** · April 18th 2008, 06:20 PM

Oh I see now, I misread the identity.

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Very Difficult Trig Identity Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

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