The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Trig And Identities

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
No, sin⁡θ2≠sin⁡θ2\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?
Reply 2
Avatar for nuodai
nuodai
17
Nope, since sin⁡(θ2)≠sin⁡θ2\sin \left( \frac{\theta}{2} \right) \ne \dfrac{\sin \theta}{2} (in general).

If it helps, make the substitution A=θ2A = \dfrac{\theta}{2}. Then you have cos⁡2A=513\cos 2A = \dfrac{5}{13} and you need to find sin⁡A\sin A. To work it out, use the double-angle formulae for cosine.
Reply 3
Avatar for Scott3142
Scott3142
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).
(edited 13 years ago)
Reply 4
Avatar for nuodai
nuodai
17
Scott3142
easy to find sin(theta) and hence sin(theta/2).

It's not particularly easy to find sin⁡θ2\sin \frac{\theta}{2} given sin⁡θ\sin \theta; conversely it's very easy to find sin⁡θ2\sin \frac{\theta}{2} given cos⁡θ\cos \theta, so that might not be the best approach.
(edited 13 years ago)
Reply 5
Avatar for Scott3142
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.
Reply 6
Avatar for nuodai
nuodai
17
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⁡θ\cos \theta like in this case an exact value for sin⁡θ2\sin \frac{\theta}{2} is expected. This is easily obtained just by using cos⁡θ=1−2sin⁡2θ2\cos \theta = 1 - 2\sin^2 \frac{\theta}{2}.
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) ?
(edited 13 years ago)
i don't understand how on earth to do it.. because there's no identity?
Reply 9
Avatar for nuodai
nuodai
17
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=1−2sin⁡2A\cos 2A = 1-2\sin^2 A

So if A=θ2A = \frac{\theta}{2}, and cos⁡θ=513\cos \theta = \frac{5}{13}, then... [you do this bit]
(edited 13 years ago)

Related discussions

Latest

Trending

Trending

Articles for you