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Math Help - Beginner! Please help with proving identities?

  1. #1
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    Beginner! Please help with proving identities?



    Hi,
    I am doing my best to understand my trig class, but I just cannot understand proving identities. I've spent the last three night searching the internet for videos and tutorials and still do not understand my own problems...

    I was able to figure out a solution to one of them by looking at a solution to a previously posted problem, but I can't get it to work for any of my others.

    I really would like to understand these!

    Problem 1:
    <br />
\frac{cos\Theta + sin\Theta}{sec\Theta + csc\Theta} = (sin\Theta)(cos\Theta)<br />

    I started with LHS. I multiplied by  (sec\Theta - csc\Theta) on both numerator and denominator and got:

    <br />
\frac{sin\Theta}{cos\Theta} - \frac{cos\Theta}{sin\Theta}

    over

    sec^2\Theta - csc^2\Theta

    And there I've lost it. Any hints? Am I even on the right track?
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Blue View Post


    Hi,
    I am doing my best to understand my trig class, but I just cannot understand proving identities. I've spent the last three night searching the internet for videos and tutorials and still do not understand my own problems...

    I was able to figure out a solution to one of them by looking at a solution to a previously posted problem, but I can't get it to work for any of my others.

    I really would like to understand these!

    Problem 1:
    <br />
\frac{cos\Theta + sin\Theta}{sec\Theta + csc\Theta} = (sin\Theta)(cos\Theta)<br />

    I started with LHS. I multiplied by  (sec\Theta - csc\Theta) on both numerator and denominator and got:

    <br />
\frac{sin\Theta}{cos\Theta} - \frac{cos\Theta}{sin\Theta}

    over

    sec^2\Theta - csc^2\Theta

    And there I've lost it. Any hints? Am I even on the right track?
    This is a different approach:

    \dfrac{\cos(\theta) + \sin( \theta)}{\sec(\theta) + \csc(\theta)} = \dfrac{\cos(\theta) + \sin( \theta)}{\dfrac1{\cos(\theta)} + \dfrac1{\sin(\theta)}} = \dfrac{\cos(\theta) + \sin( \theta)}{\dfrac{\sin(\theta)+\cos(\theta)}{\sin(\t  heta) \cdot \cos(\theta)}} = \sin(\theta) \cdot \cos(\theta)

    Now start with the RHS of the identity.
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  3. #3
    Member pflo's Avatar
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    Albuquerque, NM
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    Quote Originally Posted by Blue View Post

    I really would like to understand these!

    Problem 1:
    <br />
\frac{cos\Theta + sin\Theta}{sec\Theta + csc\Theta} = (sin\Theta)(cos\Theta)<br />

    I started with LHS. I multiplied by  (sec\Theta - csc\Theta) on both numerator and denominator and got:

    <br />
\frac{sin\Theta}{cos\Theta} - \frac{cos\Theta}{sin\Theta}

    over

    sec^2\Theta - csc^2\Theta

    And there I've lost it. Any hints? Am I even on the right track?
    Not a bad start at all. When you've got a binomial in the denominator, multiplying by the conjugate is always something to think about doing. One hint here, is that WHENEVER you have a squared trig function, think about using a Pythagorean identity. That would lead to putting in some tan^2\thetas and some cot^2\thetas in your denominator. I don't see how that would help though, because the denominator would look very similar to what you've already got there with the sec^2\thetas and csc^2\thetas. At that point, I'd say punt - which means start over and follow the idea below.

    When you can't see ANYTHING else to do, an idea that can frequently be very useful (but can also lead to lots of work) is to change everything in the equation to sin\theta and cos\theta. For this problem, that would lead to \frac{cos\theta + sin\theta}{\frac{1}{cos\theta}+\frac{1}{sin\theta}  }. You can simplify this algebraically.
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  4. #4
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    Thank you both! I can't believe I didn't see that before.
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