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Math Help - Trig Equation with varied trig functions

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    Trig Equation with varied trig functions

    Hi again all,

    sec x tan x - cos x cot x = sin x

    I've been looking at this for about 10 mins trying to figure out what first steps to take. Is there a general rule I should follow in terms of what to do first?
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    Quote Originally Posted by DannyMath View Post
    Hi again all,

    sec x tan x - cos x cot x = sin x

    I've been looking at this for about 10 mins trying to figure out what first steps to take. Is there a general rule I should follow in terms of what to do first?
    Are you sure you dont have these in your book like secx = \frac{1}{cosx}

    and  tanx = \frac{sinx}{cosx} and so on?

    You need to look at your book to know the relation between trigonometric identities.

    Show your work if you have problems
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    Quote Originally Posted by harish21 View Post
    Are you sure you dont have these in your book like secx = \frac{1}{cosx}

    and  tanx = \frac{sinx}{cosx} and so on?

    You need to look at your book to know the relation between trigonometric identities.

    Show your work if you have problems
    I know the trig identities and I went about trying to apply some but I have trouble knowing which to use and how to continue, I will write down what I have in my book that I stopped at:

    (sin x/cos^2x) - cos x (1/tanx) = sin x

    from there I tried to apply pythagorean theorem for cos^2x but I just dont have enough experience to work it out further, I did keep trying and I will keep trying but for now I am reviewing my whole term in preparation for final. I really don't mean to ask for answers, I DO want to understand these things and I've done 50 exercises just today on it.
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    Quote Originally Posted by DannyMath View Post
    I know the trig identities and I went about trying to apply some but I have trouble knowing which to use and how to continue, I will write down what I have in my book that I stopped at:

    (sin x/cos^2x) - cos x (1/tanx) = sin x

    from there I tried to apply pythagorean theorem for cos^2x but I just dont have enough experience to work it out further, I did keep trying and I will keep trying but for now I am reviewing my whole term in preparation for final. I really don't mean to ask for answers, I DO want to understand these things and I've done 50 exercises just today on it.
    how about looking at the left hand side of the equation:


    since cotx=\frac{cosx}{sinx}, you can write

    \frac{sinx}{cos^{2}x} - \frac{cos^{2}x}{sinx} = ....

    can you move ahead now? You should be able to get to the destination. Reply back if you still have problems
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    I seem to have gotten as far as:

    (sin^2 x - (1-sin^2 x)^2)/((1-sin^2 x)(sin x)) = sin x

    I got that from using the cos^2 x pythagorean identity and then trying to treat them like regular subtraction for fractions (use common denominator).

    This is the last section of the precal course and I will admit I have trouble with trig identities
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by DannyMath View Post
    I seem to have gotten as far as:

    (sin^2 x - (1-sin^2 x)^2)/((1-sin^2 x)(sin x)) = sin x

    I got that from using the cos^2 x pythagorean identity and then trying to treat them like regular subtraction for fractions (use common denominator).

    This is the last section of the precal course and I will admit I have trouble with trig identities
    Are you trying to prove the given question or are you trying to find the values of x for which the equation is true?
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    Trying to find all solutions in the interval [0, 2pi).
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by DannyMath View Post
    I seem to have gotten as far as:

    (sin^2 x - (1-sin^2 x)^2)/((1-sin^2 x)(sin x)) = sin x

    I got that from using the cos^2 x pythagorean identity and then trying to treat them like regular subtraction for fractions (use common denominator).

    This is the last section of the precal course and I will admit I have trouble with trig identities
    \frac{(sin^2 x - (1-sin^2 x)^2)}{((1-sin^2 x)(sin x))} = sinx

    (sin^2 x - (1-2sin^2 x+sin^4x) = sinx (1-sin^2 x) (sinx)

    sin^2 x - 1 + 2sin^2 x- sin^4x = sin^2x - sin^4x

    cancel the similar terms on the left side and the right side, then you can try solving for x now. Reply if you have any errors.
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    Haha this problem seems rather convoluted to me but maybe it's because I'm new to it! My answers ended up being:

    x = pi/6, 5pi/6, 7pi/6, and 11pi/6

    Hope I've come to the right solution.

    Thank you!
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by DannyMath View Post
    Haha this problem seems rather convoluted to me but maybe it's because I'm new to it! My answers ended up being:

    x = pi/6, 5pi/6, 7pi/6, and 11pi/6

    Hope I've come to the right solution.

    Thank you!
    [0, 2pi). thats what you said the interval of x was! And the answers you have gotten do not look correct.

    sin^2 x - 1 + 2sin^2 x- sin^4x = sin^2x - sin^4x

    this gives:

    -1 + 2sin^2 x = 0

    2sin^2x = 1

    sin^2x = \frac{1}{2}

    now find out what sinx would be? Then try to figure out the values of x from [0, 2\pi) for which your result is true!
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    Oh right, I forgot to use the square root right? So then the reference angle would be pi/4 rather than pi/6? If so then I've done all my homework and can start on chemistry final hehe.

    x = pi/4, 3pi/4, 5pi/4, 7pi/4

    Thanks again!
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by DannyMath View Post
    Oh right, I forgot to use the square root right? So then the reference angle would be pi/4 rather than pi/6? If so then I've done all my homework and can start on chemistry final hehe.

    x = pi/4, 3pi/4, 5pi/4, 7pi/4

    Thanks again!
    Great. I was assuming you forgot to take the square root while calculating for x. The answers look good now!

    Good luck for your Chemistry final! I absolutely hate chemistry!
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    Haha the only thing I really like about Chem is the history behind it. Cheers!
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