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(1 + cot + tan )(sin - cos ) whole upon sin^3 - cosec^3

  1. Aug 1, 2012 #1
    (1 + cot + tan )(sin - cos ) whole upon sin^3 - cosec^3 = sin^2cos^2

    prove this . ( assume theta is front of all functions ) , i dont know where to start :/
     
  2. jcsd
  3. Aug 1, 2012 #2

    HallsofIvy

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    Re: trignometry

    I spent quite a long time working on that and then realized I had miscopied it!
    And I wonder if you haven't miscopied it! Because what you have written cannot be proven- it is not true. If you take [itex]\theta= \pi/4[/itex] radians or 45 degrees, then [itex]sin(\theta)= cos(\theta)= \sqrt{2}/2[/itex] so that [itex]sin(\theta)- cos(\theta)= 0[/itex] and the numerator on the left is 0. We need to check that [itex]sin^3(\theta)- cosec^3(\theta)= 2^{3/2}/8- 8/2^{3/2}[/itex] is not 0. Since it is not the left side is 0. But the right side is [itex]sin^2(\theta)cos^2(\theta)= \frac{1}{2}\frac{1}{2}= \frac{1}{4}[/itex], NOT 0.

    Perhaps you miscopied? While I didn't finish, if the denominator were [itex]sin^3- cos^3[/itex] rather than [itex]sin^3- cosec^3[/itex], my counter-example would not work.
     
    Last edited: Aug 1, 2012
  4. Aug 1, 2012 #3
    Re: trignometry

    no , i rechecked the question , it is correct and can be solved according to the publisher ( i called customer care )
     
  5. Aug 1, 2012 #4

    Mentallic

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    Re: trignometry

    What HallsofIvy is saying is that it can't be proven because it is not true. Whether some so-called customer care said so or not, it doesn't change the fact that mathematically they're not equivalent.

    Just a couple of questions, is this what the problem says?

    Prove that
    [tex]\frac{(1+\cot \theta+\tan \theta)(\sin\theta - \cos\theta)}{\sin^3\theta-\csc^3\theta}=\sin^2\theta\cos^2\theta[/tex]

    If it is, then it can't be proven because there exist values of [itex]\theta[/itex] such that the left-hand side does not equal to the right-hand side.

    However, if the problem asked to solve rather than to prove, then we have a different question entirely. When the problem asks to solve for [itex]\theta[/itex] what it's essentially asking is to find all the value(s) of [itex]\theta[/itex] such that the left-hand side does equal the right-hand side. For example, [itex]\theta=0[/itex] satisfies the equality.
     
  6. Aug 2, 2012 #5
    Re: trignometry

    yes , that is correct , but if it cant be solved , im sorry, thank you for your time .
     
  7. Aug 2, 2012 #6

    Mentallic

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    Re: trignometry

    It can be solved, but not proven to be an identity :wink:

    You can solve [itex]x^2-2x+1=0[/itex] but you cannot prove that [itex]x^2-2x+1=0[/itex] because that implies that for every value of x, [itex]x^2-2x+1=0[/itex] which is untrue.

    You can prove that [itex]\sin^2\theta+\cos^2\theta=1[/itex] which means if you solve for [itex]\theta[/itex] you'll find that [itex]\theta[/itex] can be any value to satisfy the condition that [itex]\sin^2\theta+\cos^2\theta=1[/itex]
     
  8. Aug 2, 2012 #7

    HallsofIvy

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    Re: trignometry

    "Prove this" means "prove that this equation is correct for all x". That is not possible because it is not true as I said above. "Solve this equation" means find those particular values of x for which the equation is true. Is that what your problem asks you to do?
     
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