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Math Help - Identities help.

  1. #1
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    Identities help.

    Ok so I've been given sin(2a) = (5/13) 0 deg< a < 45 deg.

    How would I go about solving for sin.. I've been trying to use the double angel formula's and not much quite works.
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  2. #2
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    Quote Originally Posted by Chimera View Post
    How would I go about solving for...
    ... for a? Take the 'arcsin' of both sides.
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  3. #3
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    perhaps I meant to say how would I find sin of a? would I still take arcsin?
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  4. #4
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    There's another identity involving sin 2θ:
    sin 2\theta = \frac{2 tan \theta}{1 + tan^2 \theta}.

    Set sin 2θ equal to 5/13:
    \frac{5}{13} = \frac{2 tan \theta}{1 + tan^2 \theta}

    Cross multiply:
    5(1 + tan^2 \theta) = 13(2 tan \theta)
    5 + 5 tan^2 \theta = 26 tan \theta

    You have a quadratic with tan θ:
    5 tan^2 \theta - 26 tan \theta + 5 = 0

    Thank god it's factorable:
    (5 tan \theta - 1)(tan \theta - 5) = 0

    There are two possibilities for tan θ:
    5 tan \theta - 1 = 0
    tan \theta = 1/5

    tan \theta - 5 = 0
    tan \theta = 5

    Reject tan θ = 5 because you'll get an angle greater than 45. So
    tan \theta = \frac{1}{5}

    Use the Pythagorean Theorem to get the hypotenuse (opposite = 1, adjacent = 5), and you'll get
    hyp = \sqrt{26}.

    Sine is defined as opposite over hypotenuse, so
    sin \theta = \frac{1}{\sqrt{26}}.
    Last edited by eumyang; July 18th 2010 at 08:20 PM. Reason: Cleaned things up
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  5. #5
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    I wish I would have known that definition of sin2a sometime yesterday...

    If that particular definition derived from any others? the only formula for sin(2a) listed in my book is sin(2a) = 2cos(x)sin(x)...

    This particular chapter seems to be horribly written.. many of the questions we are asked don't seemed to be covered in the book.. or require different methods than what we have been given..

    I greatly appreciate the help.
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  6. #6
    Senior Member eumyang's Avatar
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    I wasn't taught the other formula for sin 2θ either with I was in school. But it's not hard to derive it from the one we're familiar with:
    \begin{aligned}<br />
sin 2\theta &= 2 sin \theta cos \theta \\<br />
&= \frac{2 sin \theta}{sec \theta} \\<br />
&= \frac{2 sin \theta}{sec \theta} \cdot \frac{sec \theta}{sec \theta} \\<br />
&= \frac{2 sin \theta \cdot \frac{1}{cos \theta}}{sec^2 \theta} \\<br />
&= \frac{2 tan \theta}{1 + tan^2 \theta}<br />
\end{aligned}
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  7. #7
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    that makes sense. This particular topic is the first I've had trouble on this semester in math X.x. So many different ways to go about it all.

    Thank you .
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  8. #8
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    My suggestion would still work (great deal easier); take the arcsin of both sides and then the sin of both sides:

    If \sin(2a) = \frac{5}{13}, then 2a = \arcsin\frac{5}{13} \Rightarrow a = \frac{\arcsin\left(\frac{5}{13}\right)}{2} \Rightarrow \sin{a} = \sin\left({\frac{\arcsin\left(\frac{5}{13}\right)}  {2}\right) =\frac{1}{\sqrt{26}}}.
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  9. #9
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    Quote Originally Posted by Chimera View Post
    If that particular definition derived from any others? the only formula for sin(2a) listed in my book is sin(2a) = 2cos(a)sin(a)...
    This works as well, but it requires you to find \cos{a} so as to find \sin{a}. Now, \sin{a} can be found in exactly
    the same as \cos{a}, so why not just find \sin{a} instead of finding \cos{a} so as to find \sin{a}? In fact, finding
    \cos{a} involves bit more of work than finding \sin{a}, as you will need to find the missing side of the triangle.
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  10. #10
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    I am not having trouble with verifying identities x.x

    in particular cos2x+cos2y
    sinx + cosy

    all equal to 2cosy- 2sinx

    my first instinct was to convert the top cosx and y on the fraction to something different but, it doesn't seem to quite pan out..
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  11. #11
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    sin(2a) = 2sin(a)cos(a) = 5/13

    [Math]4sin^2(a)cos^2(a) = 25/169[/tex]

    [Math]4sin^2(a)[1- sin^2(a)] = 25/169[/tex]

    [Math]4sin^2(a) - 4sin^4(a) = 25/169[/tex]

    Solve the quadratic and find sin(a)
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