Solving 2 Trigo Identities

**liukawa** · November 13th 2010, 11:14 PM

hi guys !

have this 2 trigo identities which i am currently stuck with.

Appreciate if anyone can enlighten me thanks !

**Educated** · November 13th 2010, 11:30 PM

Originally Posted by liukawa

hi guys !

have this 2 trigo identities which i am currently stuck with.

Appreciate if anyone can enlighten me thanks !Attachment 19697

I'm sorry but I get the message "Invalid Attachment specified." when I click on the attachment.
Maybe you could try re-uploading it?

**liukawa** · November 14th 2010, 12:28 AM

hey dude.

i reposted !

thanks for informing me about the glitch !

**Soroban** · November 14th 2010, 04:22 AM

Hello, liukawa!

$\text{Prove that: }\;2\csc^2\!2x - \csc^2\!x \;=\;-2\cot2x\csc2x$

The left side is: . $\dfrac{2}{\sin^2\!2x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{(2\sin x\cos x)^2} - \dfrac{1}{\sin^2\!x}$

. . . $=\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{4\cos^2\!x}{4\sin^2\!x\cos^2\!x}$

. . . $\displaystyle =\;\;\frac{2-4\cos^2\!x}{4\sin^2\!x\cos^2\!x} \;\;=\;\;\frac{2 - 4\left(\frac{1+\cos2x}{2}\right)}{\sin^22x} \;\;=\;\;\frac{2 - 2 - 2\cos2x}{\sin^22x}$

. . . $\displaystyle =\; \frac{-2\cos2x}{\sin^22x} \;=\; -2\cdot\frac{\cos2x}{\sin2x}\cdot\frac{1}{\sin2x} \;=\;-2\cot2x\csc2x$

**liukawa** · November 14th 2010, 04:35 AM

Originally Posted by Soroban

Hello, liukawa!

The left side is: . $\dfrac{2}{\sin^2\!2x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{(2\sin x\cos x)^2} - \dfrac{1}{\sin^2\!x}$

. . . $=\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{4\cos^2\!x}{4\sin^2\!x\cos^2\!x}$

. . . $\displaystyle =\;\;\frac{2-4\cos^2\!x}{4\sin^2\!x\cos^2\!x} \;\;=\;\;\frac{2 - 4\left(\frac{1+\cos2x}{2}\right)}{\sin^22x} \;\;=\;\;\frac{2 - 2 - 2\cos2x}{\sin^22x}$

. . . $\displaystyle =\; \frac{-2\cos2x}{\sin^22x} \;=\; -2\cdot\frac{\cos2x}{\sin2x}\cdot\frac{1}{\sin2x} \;=\;-2\cot2x\csc2x$

Thanks dude !

You are awesome !

Any tips on how to solve the 2nd trigo identity?

Have been cracking my head whole day long and havent got a single clue !

Appreciate your help greatly thanks !

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