trig identities!!!help!!!

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April 30th 2006, 05:43 AM
awesometrackgirl

trig identities!!!help!!!

i need major help with some problems, i want to be able to figure them out on my own, so please don't give me the answers

okay i got so far, but i don't think i did it right
here's the problem

(cosx)(sec3x+cotx)=tan2x\1-cos2x+sinx\sec2-1

(sinx\cosx)(sinx\cosx)/sin2x+sin\tan

then i crossed out the sins but i don't think i did it right, please help!!
April 30th 2006, 06:58 AM
CaptainBlack

Quote:

Originally Posted by awesometrackgirl

i need major help with some problems, i want to be able to figure them out on my own, so please don't give me the answers

okay i got so far, but i don't think i did it right
here's the problem

(cosx)(sec3x+cotx)=tan2x\1-cos2x+sinx\sec2-1

(sinx\cosx)(sinx\cosx)/sin2x+sin\tan

then i crossed out the sins but i don't think i did it right, please help!!

It looks like you are asking for help in proving the following trig identity:

cos(x)[sec^3(x)+cot(x)]=tan^2(x)/(1-cos^2(x)) + sin(x)/(sec^2(x)-1)

which is how we would normaly write this is plain ASCII.

It is easier to see what this is if we rewrite it in TeX:

$ \cos(x)[\sec^3(x)+\cot(x)]=\frac{\tan^2(x)}{1-\cos^2(x)} + \frac{\sin(x)}{\sec^2(x)-1} $

RonL
April 30th 2006, 07:15 AM
CaptainBlack

We wish to prove the trig identity:

$ \cos(x)[\sec^3(x)+\cot(x)]=\frac{\tan^2(x)}{1-\cos^2(x)} + \frac{\sin(x)}{\sec^2(x)-1} $

Focus on the RHS, look at the first term:

$ \frac{\tan^2 (x)}{1-\cos^2(x)} $

use $\sin^2(x)=1-cos^2(x)$ to simplify the denominator, then try to
rearrange what you have to be the first term on the LHS after the bracket is
expanded, i.e. the first term of:

$ \cos(x)\sec^3(x)+\cos(x)\cot(x) $

Now repeat the process with the second term on the RHS (and show it is
equal to the second term in the expression above.

RonL