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Math Help - Trig problem

  1. #1
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    Trig problem

    Hi. Got a maths trig question.
    For a angle of less than 90 degrees, sinx=3cosx. Show and decide the value of sin2x.

    I just cant bring myself to figure this one out. Helps or pointers would be very appreciated
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  2. #2
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    Quote Originally Posted by Burger king View Post
    Hi. Got a maths trig question.
    For a angle of less than 90 degrees, sinx=3cosx. Show and decide the value of sin2x.

    I just cant bring myself to figure this one out. Helps or pointers would be very appreciated
    \sin(2x) = 2\sin{x}\cos{x}

    since \sin{x} = 3\cos{x} ...

     <br />
\sin(2x) = 2(3\cos{x})\cos{x} = 6\cos^2{x}<br />
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  3. #3
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    Hello, Burger king!

    I can't imagine what "show and decide" means.


    For an angle x < 90^o\!:\;\;\sin x\:=\:3\cos x.
    Show and decide the value of \sin2x.

    We have: . \sin x \:=\:3\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:=\:3 \quad\Rightarrow\quad \tan x \:=\:3
    Hence: . \tan x \:=\:\frac{3}{1} \:=\:\frac{opp}{adj}

    So x is an angle in a right triangle with: opp = 3,\;adj = 1
    . . Pythagorus says: . hyp = \sqrt{10}
    Hence: . \sin x = \tfrac{3}{\sqrt{10}},\;\;\cos x = \tfrac{1}{\sqrt{10}}

    Therefore: . \sin2x \:=\:2\sin x\cos x \:=\:2\left(\frac{3}{\sqrt{10}}\right)\left(\frac{  1}{\sqrt{10}}\right) \:=\:\frac{3}{5}

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