Identifying trig identities. need help

[Freshgeardude]

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

 

[Epic Fail]

42

 

[TuxDave]

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.

 

[Freshgeardude]

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.

i am still trying though

 

[totalnoob]

42

indeed, here is proof of the trig identity you seek..

http://lmgtfy.com/?q=What+is+the+answer+to+life,+the+universe,+and+everything?

 

[TuxDave]

i am still trying though

Let a = sin^2(x)
Let b = cos^2(x)

sin^4(x) - cos^4(x) = a^2 - b^2 = (a+b)(a-b)

Did you get at least this far?

 

[Hacp]

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

I hope you're still in high school.

 

[DrPizza]

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.

 

[benzylic]

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.