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anonymous

  • one year ago

Trig / Pre Cal/identities Am I on the right path? Please do not give the answer. Here is the problem \( \frac{(\sin x- \cos x)*2}{cos} = \sec x - \cos x\) I am working on the left side and need to prove it equals the right. This is what I have done so far \( \frac{(\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x\) Am I going about this the right way because I don't see where to go from here. Thank you.

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  1. anonymous
    • one year ago
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    It is suppose to be Trig / Pre Cal/identities Am I on the right path? Please do not give the answer. Here is the problem \( \frac{(\sin x- \cos x)^2}{cos} = \sec x - \cos x\) I am working on the left side and need to prove it equals the right. This is what I have done so far \( \frac{(\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x\) Am I going about this the right way because I don't see where to go from here. Thank you.

  2. anonymous
    • one year ago
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    This is where I am at. Am I on the right path and where do I go from here because I do not see anything. \( \frac{\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x\)

  3. anonymous
    • one year ago
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    Sorry

  4. anonymous
    • one year ago
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    Made a mistake

  5. Nnesha
    • one year ago
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    hmm well....... i guess both sides are not equal

  6. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nixy Made a mistake \(\color{blue}{\text{End of Quote}}\) what ?? :-)

  7. anonymous
    • one year ago
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    It is suppose to be \( \frac{(\sin x- \cos x)^2}{cos} = \sec x - 2\sin x \) And I am at \( \frac{\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - 2\sin x \)

  8. anonymous
    • one year ago
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    I was typing it in too fast and made a boo boo. Sorry about that.

  9. anonymous
    • one year ago
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    Try sin² x = 1 - cos² x

  10. anonymous
    • one year ago
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    So the numerator becomes \[1 - cos^2 x - 2 (sin x)(cos x) + cos^2 x\]

  11. Nnesha
    • one year ago
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    that's right or just put 1 sin^2x+cos^2 =1 you will get the same answer :-)

  12. anonymous
    • one year ago
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    That is it peachpi.

  13. anonymous
    • one year ago
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    I did that a min ago and it did not work but re looked at it and I see where I made my mistake. Thank you all

  14. anonymous
    • one year ago
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    I think I over think these things at time. One minute I am rolling through them and then I get snagged by the simplest thing lol Thank you all

  15. Nnesha
    • one year ago
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    haha great work!!! :-)

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