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    Given that x is an acute angle and sin x =12/13, use the identity cos^2 x + sin^2 x= 1 to show that 13cos x =5.
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    (Original post by man111111)
    Given that x is an acute angle and sin x =12/13, use the identity cos^2 x + sin^2 x= 1 to show that 13cos x =5.
    Cool, let's work through this.

    Do you know what an acute angle is? An angle between 0 and 90 degrees.

    So,  sin(0<x<90)=\frac{12}{13}

     (\sin x)^2 = \frac{144}{169}

     1 - (\cos x)^2 = \frac{144}{169}

    Can you finish?
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    (Original post by Apachai Hopachai)
    Cool, let's work through this.

    Do you know what an acute angle is? An angle between 0 and 90 degrees.

    So,  sin(0<x<90)=\frac{12}{13}

     (\sin x)^2 = \frac{144}{169}

     1 - (\cos x)^2 = \frac{144}{169}

    Can you finish?

    Hi, thanks for replying. I don't know how to finish it?
    Is (cos x)^2 the same as cos^2 x
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    (Original post by man111111)
    Hi, thanks for replying. I don't know how to finish it?
    Is (cos x)^2 the same as cos^2 x
    Yes it is. NOW do you know how to finish?
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    (Original post by RDKGames)
    Yes it is. NOW do you know how to finish?
    Hi thanks for replying. I don't know how to finish it.
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    (Original post by man111111)
    Hi thanks for replying. I don't know how to finish it.
    Okay.....

    If you say Y=\cos(x) then your equation becomes 1-Y^2=\frac{144}{169} and you want to solve for Y.

    Is this clearer...?
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    (Original post by man111111)
    Hi thanks for replying. I don't know how to finish it.
    This is not food, you can't just throw it in the bin if you can't finish it.

    FINISH IT
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    (Original post by RDKGames)
    Okay.....

    If you say Y=\cos(x) then your equation becomes 1-Y^2=\frac{144}{169} and you want to solve for Y.

    Is this clearer...?
    Thank you for helping me.
 
 
 
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