Need help with Trigonometric Identity questions.

**Xenophobe** · April 18th 2009, 05:59 PM

Hey, need some help with two questions (may be adding more to this thread later). I am finding these problems rather difficult, so an explanation of each step needed would be EXTREMELY HELPFUL!
1. $(secx - cosx)(cscx - sin x) = \frac{tanx}{1+tan^2x}$
My attempt:
LHS
= $(\frac{1}{cosx} - cosx)(\frac{1}{sinx} - sinx)$ *changed secx to 1/cosx, changed cscx to 1/sinx
= $(\frac{sin^2x + cos^2x}{cosx} - cosx)(\frac{sin^2x + cos^2x}{sinx} - sinx)$ *replaced "1" with pythagrean
This is where I am stuck as to what to do.

2. $sec^6x - tan^6x = 1 + 3tan^2x sec^2x$
My attempt:
LHS
= $(\frac{1}{cos^6x}) - (\frac{sin^6x}{cos^6x})$ *changed tanx to sinx/cosx, changed secx into 1/cosx
= $(\frac {sin^2x + cos^2x} {cos^6x}) - (\frac{sin^6x}{cos^6x})$ * Replaced "1" with pythagrean
Unknown what to do from here.

EDIT: Okay, I just read the rules. My apologies for all with the title involving "help" I should have been more specific. Unknown if possible to change title or not, I do not see a way right now.
What I did in the edit:
Showed my attempts and tried to explain what I was thinking for each step.

EDIT2: K, everything is in LaTeX! Took about 10 minutes extra, but WOW, looks a lot neater and easier to understand! Really glad I read the rules.

**Reckoner** · April 18th 2009, 07:44 PM

I did these through sheer brute force. You may be able to clean it up a bit.

$(\sec x-\cos x)(\csc x-\sin x)=\frac{\tan x}{1+\tan^2x}$

Expand the left side:

$(\sec x-\cos x)(\csc x-\sin x)$

$=\sec x\csc x-\sin x\sec x-\cos x\csc x+\sin x\cos x$

$=\sec x\csc x-\tan x-\cot x+\sin x\cos x$

$=\frac1{\sin x\cos x}+\sin x\cos x-\tan x-\cot x$

$=\frac{1+\sin^2x\cos^2x-\sin^2x-\cos^2x}{\sin x\cos x}$

$=\frac{1+\sin^2x\cos^2x-1}{\sin x\cos x}=\frac{\sin^2x\cos^2x}{\sin x\cos x}$

$=\sin x\cos x$

$=\frac{\sin x}{\sec x}=\frac{\sin x\sec x}{\sec^2x}$

and you can hopefully finish this last step.

$\sec^6x-\tan^6x=1+3\tan^2x\sec^2x$

$\sec^6x-\tan^6x$

$=\frac{1-\sin^6x}{\cos^6x}$

$=\frac{\left(1-\sin^2x\right)\!\left(1+\sin^2x+\sin^4x\right)}{\c os^6x}$

$=\frac{\cos^2x\left(1+\sin^2x+\sin^4x\right)}{\cos ^6x}$

$=\frac{1+\sin^2x+\sin^4x}{\cos^4x}$

$=\sec^4x+\tan^2x\sec^2x+\tan^4x$

$=\sec^4x+\left(\sec^2x-1\right)\sec^2x+\tan^4x$

$=2\sec^4x-\sec^2x+\tan^4x$

$=2\sec^4x-\sec^2x+\left(\sec^2x-1\right)^2$

$=3\sec^4x-3\sec^2x+1$

$=3\sec^2x\left(\sec^2x-1\right)+1$

$=1+3\tan^2x\sec^2x$

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