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Urgent Help Needed, C3 maths, Trig Functions!!

Ok, im really stuck on a certain question, and have been through the whole book and internet and cannot find anything to help me, so hopefully you can?

The question is:

Show that the equation
Sin( x + 30° ) = 2 cos ( x + 60° )
Can be written in the form 3(root)3 sin x = 1 cos x

Thanks
Reply 1
cosθsin(90θ) \cos \theta \equiv \sin (90-\theta ) .
Reply 2
Original post by B_9710
cosθsin(90θ) \cos \theta \equiv \sin (90-\theta ) .


But don't you have to keep both the cos and the sin as they are both in the final answer?
Just expand and rearrange to get the final expression.
(edited 7 years ago)
Reply 4
Original post by BAMItzCallum
But don't you have to keep both the cos and the sin as they are both in the final answer?


I will look at it properly. Give me a minute.
Reply 5
Thanks :smile:
Original post by BAMItzCallum
Ok, im really stuck on a certain question, and have been through the whole book and internet and cannot find anything to help me, so hopefully you can?

The question is:

Show that the equation
Sin( x + 30° ) = 2 cos ( x + 60° )
Can be written in the form 3(root)3 sin x = 1 cos x

Thanks


Expand Sin (A+B) = sin A cos B + cos A sin B and cos (A+B) = cosA cos B - sin A sin B and use sin 60 = root3/2 = cos 30 and sin 30= 1/2 = cos 60. Then rearrange to get final expression.
Reply 7
Original post by tangotangopapa2
Just expand and rearrange to get the final expression.


I tried that, and came out with (root3)/2 sinx + 1/2 cosx = 1 cox - (root3) sinx
Which is completely wrong
That is not wrong take cos x terms to one side and sin x terms to other side. you get 3 root 3 sinx /2 = cos x /2. Cancel 2 to get final expression

Original post by BAMItzCallum
I tried that, and came out with (root3)/2 sinx + 1/2 cosx = 1 cox - (root3) sinx
Which is completely wrong
Reply 9
Original post by tangotangopapa2
That is not wrong take cos x terms to one side and sin x terms to other side. you get 3 root 3 sinx /2 = cos x /2. Cancel 2 to get final expression


if i put them on the same side though, i get:
(root3)/2 sinx + (root3)sinx = 1cosx + 1/2 cosx
:/
root3/2 + root3 is 3root3/2
You should get 1 cos x - 1/2 cos x on RHS.

Original post by BAMItzCallum
if i put them on the same side though, i get:
(root3)/2 sinx + (root3)sinx = 1cosx + 1/2 cosx
:/
Original post by tangotangopapa2
root3/2 + root3 is 3root3/2
You should get 1 cos x - 1/2 cos x on RHS.


Derp. Thanks so much!
Original post by BAMItzCallum
Derp. Thanks so much!


You're very welcome.

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