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Identifying trig identities. need help

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Step 1) Never use WolframAlpha when trying to do your homework. It'll hurt you in the long term because reading the answer is infinitely worse than trial and error as you try to figure out the right steps.

Step 2) I can see the answer in my head. Sometimes it's not just trig identities but you should also use some algebraic identities. a^2-b^2 for instance.
 
TuxDave said:
Step 1) Never use WolframAlpha when trying to do your homework. It'll hurt you in the long term because reading the answer is infinitely worse than trial and error as you try to figure out the right steps.

thanks lol I know, but I have spent over 20 minutes on this problem and I thought linking wolfram so people can see it the way I see it would help.

Step 2) I can see the answer in my head. Sometimes it's not just trig identities but you should also use some algebraic identities. a^2-b^2 for instance.
Click to expand...


i am still trying though
 
Freshgeardude said:
trying to solve this.


(sin^4(x)-cos^4(x))/(sin^2(x)) = 1-cot^2(x)

http://www.wolframalpha.com/input/?i=%28sin^4x+-+cos^4x%29%2F%28sin^2x%29+%3D+1-cot^2x

I cant seem to get it. I know the trig identities, but every time I get lost.

need some help please.



oh yea.. this is homework, so in before trolls
Click to expand...

I hope you're still in high school.

sin4.GIF
 
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Still stuck? If so, note that after factoring on the left (difference of two perfect squares), one of those factors is the easiest pythagorean trig identity. Since you have a fraction on the left (one term), and two terms on the right, break your fraction into two terms. Done.
 
factor the numerator into (sin^2x-cos^2x)(sin^2x+cos^2x)

Then you can use the sin^2x + cos^2x =1 to eliminate the right set of parenthesis. You should be able to get it after that.
 
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