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Math Help - trigonometry question 2

  1. #1
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    Exclamation trigonometry question 2

    If x is an acute angle and cosx=4/5, find the values of sin2x, sin3x, tan3x.
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  2. #2
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    If \cos{x} = \frac{4}{5}, then x = cos^{-1}\left( \frac{4}{5}\right) = 36.9^o (acute angle).

    Simply substitute this x value into the other equations:

    \sin{(2 \times 36.9^o)} = ?

    \sin{(3 \times 36.9^o)} = ?

    \tan{(3 \times 36.9^o)} = ?
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  3. #3
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    If you need an exact answer, you can use the definition of cos using a triangle.

    trigonometry question 2-trigo.jpg

    Pythagoras' theorem can help you find sin(x). Then you can find sin(2x) and cos(2x) using the double angle formulas and then use sin(3x) = sin(x+2x) and cos(3x) = cos(2x+x)
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  4. #4
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    Quote Originally Posted by math123456 View Post
    if x is an acute angle and cos x=4/5, find the exact values of sin2a, sin3a and tan3a.
    cos(x) = adjacent/hypotenuse.

    So adjacent = 4
    Hypotenuse = 5.

    You can use pythagorus to work out the opposite side, and the use that to find sin(x) = opposite/hypotenuse.

     sin2x = 2sin(x)cos(x)

     sin(3x) = sin(2x+x) = sin(2x)cos(x) + sin(x)cos(2x)

    = 2sin(x)cos(x)cos(x)+sin(x)(cos^2(x)-sin^2(x))

     tan(3x) = \frac{sin(3x)}{cos(3x)} = \frac{sin(2x+x)}{cos(2x+x)}=\frac{sin(2x)cos(x) + sin(x)cos(2x)}{cos(2x)cos(x)-sin(2x)sin(x)}

     =\frac{2sin(x)cos(x)cos(x)+sin(x)(cos^2(x)-sin^2(x))}{(cos^2(x)-sin^2(x))cos(x)-2sin(x)cos(x)sin(x))}
    Last edited by mr fantastic; January 1st 2009 at 08:47 PM. Reason: No edit - just flagging this reply as having been moved from a double post
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by math123456 View Post
    if x is an acute angle and cos x=4/5, find the exact values of sin2a, sin3a and tan3a.
    Ok since \cos{x} = \frac{4}{5} is an acute ww can concur that sin is positive as well

    Using pythagorean theorem we see that \sin{x} = \frac{3}{5}

    Now \sin{2x} = 2\sin{x}\cos{x}

    Now just plug in and solve
    Last edited by mr fantastic; January 1st 2009 at 08:47 PM. Reason: No edit - just flagging this reply as having been moved from a double post
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