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    if theta is an acute angle such that cos theta = 5/13 then find the exact value of sin (theta/2)

    can you find sin theta and then divide the answer by two? i can't get to the right answer! hellppppp
    thanks
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    No, \sin \frac{\theta}{2} \neq \frac{\sin \theta}{2}.

    Do you know any formulae that relate trig functions where one of the angles involved is double the other?
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    Nope, since \sin \left( \frac{\theta}{2} \right) \ne \dfrac{\sin \theta}{2} (in general).

    If it helps, make the substitution A = \dfrac{\theta}{2}. Then you have \cos 2A = \dfrac{5}{13} and you need to find \sin A. To work it out, use the double-angle formulae for cosine.
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    You could draw a right angled triangle, and put the values in that allow cos(theta) = 5/13, as it happens, its a nice pythagorean triple as well, so easy to find sin(theta) and hence sin(theta/2).
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    (Original post by Scott3142)
    easy to find sin(theta) and hence sin(theta/2).
    It's not particularly easy to find \sin \frac{\theta}{2} given \sin \theta; conversely it's very easy to find \sin \frac{\theta}{2} given \cos \theta, so that might not be the best approach.
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    I agree, however given that theta is an acute angle, that is to say only one value, is it not valid to just say that sin(theta)=12/13, then theta = arcsin(12/13), theta/2=[arcsin(12/13)]/2, sin(theta/2) = sin(arcsin(12/13)/2). Not particularly pretty, but a valid answer nonetheless, and no need to mess about with double angle formulae.
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    (Original post by Scott3142)
    I agree, however given that theta is an acute angle, that is to say only one value, is it not valid to just say that sin(theta)=12/13, then theta = arcsin(12/13), theta/2=[arcsin(12/13)]/2, sin(theta/2) = sin(arcsin(12/13)/2). Not particularly pretty, but a valid answer nonetheless, and no need to mess about with double angle formulae.
    It would give you an answer, but often when you're given an exact value for \cos \theta like in this case an exact value for \sin \frac{\theta}{2} is expected. This is easily obtained just by using \cos \theta = 1 - 2\sin^2 \frac{\theta}{2}.
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    (Original post by Scott3142)
    I agree, however given that theta is an acute angle, that is to say only one value, is it not valid to just say that sin(theta)=12/13, then theta = arcsin(12/13), theta/2=[arcsin(12/13)]/2, sin(theta/2) = sin(arcsin(12/13)/2). Not particularly pretty, but a valid answer nonetheless, and no need to mess about with double angle formulae.
    The question wants an exact answer. How are you intending to evaluate sin(arcsin(12/13)/2) ?
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    i don't understand how on earth to do it.. because there's no identity?
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    (Original post by sophiestanmoor110)
    i don't understand how on earth to do it.. because there's no identity?
    Read the posts more carefully... there's more than enough information.

    To make it more obvious, the identity you need to use is \cos 2A = 1-2\sin^2 A

    So if A = \frac{\theta}{2}, and \cos \theta = \frac{5}{13}, then... [you do this bit]
 
 
 
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