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Rewrite in terms of sinØ and cosØ:

  1. Jun 5, 2005 #1
    Rewrite in terms of sinØ and cosØ:
    Code (Text):

    --------  +1

    Code (Text):
        tan x
    csc x + cot x
    A friend of mine and I have been struggling over these type of problems for the past few hours, unable to find any logical and simple way to solve them. We have the answers already thanks to our handy answer key, but we need to know HOW to get the answers. If someone would be so kind enough as to walk us through these, that'd be great. We have a test tomorrow, and need to know quick ways to arrive at an answer while taking the final, so be accurate, heh.​

    Last edited: Jun 5, 2005
  2. jcsd
  3. Jun 5, 2005 #2
    Oh; and for your convenience, here are the answers:

    Rewrite in terms of sinØ and cosØ:

    Code (Text):

    Code (Text):
     sec x - 1
    Thanks again.​
  4. Jun 5, 2005 #3
    what about writing everything in terms of sines & cosines? there are also only a few identities you can use. what have you done so far?

    edit: ok i just did the 1st one by doing what i said, writing cot as cos/sin, getting a common denominator & the identity cos^2 + sin^2 = 1, cancelling (1 + sin) to get the correct answer.

    edit again: for the 2nd one write as cos & sin everywhere, simplify, write (1-cos^2) as (1+cos)(1-cos) to be able to cancel (1+cos) & get the correct answer
    Last edited: Jun 5, 2005
  5. Jun 5, 2005 #4
    Rewrite in terms of sinØ and cosØ:

    For this problem we did the basic stuff: we rewrote it in terms of sin and cos and started doing a little dance that really didn't help us. Honestly, we tried a few identities here and there, but all of them ended up making the problem more elaborate, which isn't what we want.​


    As for this problem, it was the same idea. We rewrote it in terms of sin and cos, and attempted to apply some identities and work for some cancelation. Sadly, it'd always end up making it more elaborate than before, which is obiouvsly not the goal. As for specific identities, I'm not sure. We tried a lot of stuff and it'd take an extensive and pointless post to recollect what they all were.​

    Thanks again for any help, and sorry that I couldn't be of any, heh.
  6. Jun 5, 2005 #5
    Could you please elaborate? Certain points you make are inconsistent with the equation and it seems like you're leaving holes to inturpret. Thanks for your concern, though - any help is greatly appreciated.
  7. Jun 5, 2005 #6
    silly me i should have just added a couple new replies rather than edits :yuck: just look up instead
  8. Jun 5, 2005 #7
    Thanks for the tips on the common denominator - we got the first one. We'll try your tactic on the second one and see if we can't manage a similar result. Thanks again :!!)

    Edit: We got them both. w00t!
    Last edited: Jun 5, 2005
  9. Jun 5, 2005 #8
    ok fine here's the 1st one

    [tex]\frac{cot^2\theta}{csc\theta + 1} + 1 = \frac{\frac{cos^2\theta}{sin^2\theta}}{\frac{1}{sin\theta + 1}}} + 1[/tex]

    put the sin^2 in the denominator:

    [tex]\frac{cos^2\theta}{sin^2\theta(\frac{1}{sin\theta}+1)} + 1 = \frac{cos^2\theta}{sin\theta + sin^2\theta} + 1[/tex]

    common denominator:

    [tex]\frac{cos^2\theta + sin^2\theta + sin\theta}{sin^2\theta + sin\theta}[/tex]

    use an identity:

    [tex]\frac{1 + sin\theta}{sin\theta + sin^2\theta} = \frac{1 + sin\theta}{sin\theta(1 + sin\theta)}[/tex]

    can you do the rest now? :wink:

    edit: lol i guess i didn't have to do that after all :rolleyes:
  10. Jun 6, 2005 #9
    [tex]\frac{cot^2\theta}{csc\theta + 1} + 1 = \frac{\frac{cos^2\theta}{sin^2\theta}}{\frac{1}{si n\theta + 1}}} + 1[/tex]

    this line should be

    [tex]\frac{cot^2\theta}{csc\theta + 1} + 1 = \frac{\frac{cos^2\theta}{sin^2\theta}}{\frac{1}{sin\theta} + 1}} + 1[/tex] or [tex]\frac{cot^2\theta}{csc\theta + 1} + 1 = \frac{\frac{cos^2\theta}{sin^2\theta}}{\frac{1 + \sin \theta}{si n\theta}}} + 1[/tex]
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