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Analytically showing that equations represent trigonometric identity statements pre calculus help?

I understand half the equations for homework and can do them easily, but then the other half just seem like they can’t be solved.

For example, (cscx+cotx)/(tanx+sinx)=cotxcscx. I can break that down to (1/sinx + cosx/sinx)/(sinx/cosx + sin)=cosx/sin^2x. But what on earth do I do from here?

Or, for another example, cotxsinx=cos^3x+cosxsin^2x. The left easily becomes cosx, I know. But what can I do with the right? I don’t see any substitutions that make anything work. I’m not sure what I’m missing. Some problems I can easily solve, but then ones like these it seems like they’re not solveable, like they’re not actually trigonometric identities, but that’s of course not an option. The way it’s worded clearly implies that they all are.

I missed a few days of class from being sick and my teacher is really bad at explaining things, she basically writes a problem on the board and then writes the answer with little to no explanation on how she’s gotten there. I’ve been trying to figure this out on my own, some of it I have, but other bits, like this, I have not.

I have a test tomorrow and need to figure this stuff out before then.

3 Answers

Relevance
  • ?
    Lv 7
    2 years ago

    Just a couple more steps...

    You have cosx/sin^2x which is (cosx/sinx) * (1/sinx) which is cotx * cscx.

  • alex
    Lv 7
    2 years ago

    cos^3x+cosxsin^2x=cosx(cos^2x+sin^2x)= cosx(1)=cosx

    Identity

    cos^2x+sin^2x =1

  • ?
    Lv 4
    2 years ago

    You are on right track. Here are the steps I use to prove these identities:

    1. Express everything in terms of cos x and sin x. (You are clearly doing this.)

    2. Look for any identities that you already know.

    3. Use algebra to rearrange the equation, e.g. factoring, finding common denominators, etc.

    Your first example:

    (1/sinx + cosx/sinx)/(sinx/cosx + sinx)=cosx/sin²x

    Algebra lets us "cross multiply" which gives:

    sin²x(1/sinx + cosx/sinx)=cosx(sinx/cosx + sinx) or

    sinx + sinxcosx = sinx + sinxcosx and you are done.

    Your second example:

    Remember the identity sin²x + cos²x = 1

    or equivalently sin²x=1-cos²x

    You reached the point where

    cosx = cos³x+cosxsin²x now substitute for sin²x

    cosx = cos³x+cosx(1-cos²x)

    cosx = cos³x+cosx-cos³x

    cosx = cosx

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