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Math Help - help with trig identities!

  1. #1
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    help with trig identities!

    1. solve : (sinX)(cosX)-1.2cosx = 0

    2. find identity or counter example : csc(3pi/4 + A) = -secA

    3. find identity or counter example : cosX = (cscXtanX)/(1+tan^2x)

    4. find identity or counter example : (sinX)cosx = tanx

    5. find identity or counter example : cos^2x = (sin^2x+cos^2x)/tan^2x + 1
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  2. #2
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    Quote Originally Posted by Joker View Post
    1. solve : (sinX)(cosX)-1.2cosx = 0
    \sin{x}\cos{x}-1.2\cos{x} = 0 \Rightarrow \cos{x}\left(\sin{x}-1.2\right) = 0 \Rightarrow \cos{x} = 0, or \sin{x} = 1.2.

    2. find identity or counter example : csc(3pi/4 + A) = -secA
    Take A = 0.

    3. find identity or counter example : cosX = (cscXtanX)/(1+tan^2x)
    \displaystyle \frac{\csc{x}\tan{x}}{1+\tan^2{x}} = \frac{1}{\cos{x}\left(1+\frac{\sin^2{x}}{\cos^2{x}  }\right)}  \displaystyle  = \frac{1}{\cos{x}\left(1+\frac{1-\cos^2{x}}{\cos^2{x}}\right)} = \frac{1}{\cos{x}\left(\frac{\cos^2{x}+1-\cos^2{x}}{\cos^2{x}}\right)} = \frac{1}{\cos{x}\left(\frac{1}{\cos^2{x}\right)}} = \cos{x}

    4. find identity or counter example : (sinx)cosx = tanx
    Oh, come on! Think about it for a bit!

    5. find identity or counter example : cos^2x = (sin^2x+cos^2x)/tan^2x + 1
    \displaystyle \frac{\sin^2{x}+\cos^2{x}}{\tan^2{x}+1} = \frac{1}{\tan^2{x}+1} = \frac{1}{\frac{\sin^2{x}}{\cos^2{x}}+1} = \frac{1}{\frac{1-\cos^2{x}}{\cos^2{x}}+1} = \cdots
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  3. #3
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    thanks for the two posts. this should help a lot! the first one was suppose to be 1/2cosx lol but i think i can handle it. Thanks Again
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  4. #4
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    Quote Originally Posted by Joker View Post
    thanks for the two posts. this should help a lot! the first one was suppose to be 1/2cosx lol but i think i can handle it. Thanks Again
    \displaystyle \sin{x}\cos{x} + \frac{1}{2}\cos{x} = 0

    \displaystyle \cos{x}\left(\sin{x} +\frac{1}{2}\right)=0

    \displaystyle \cos{x} = 0 or \displaystyle \sin{x} + \frac{1}{2} = 0.


    Case 1: \displaystyle \cos{x} = 0

    \displaystyle x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n where \displaystyle n \in \mathbf{Z}.


    Case 2: \displaystyle \sin{x} + \frac{1}{2} = 0

    \displaystyle \sin{x} = -\frac{1}{2}

    \displaystyle x = \left\{\pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} + 2\pi n where \displaystyle n \in \mathbf{Z}

    \displaystyle x = \left\{\frac{7\pi}{6}, \frac{11\pi}{6}\right\} + 2\pi n.


    So the solution is \displaystyle x = \left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\} + 2\pi n.
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