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Core Maths 4: Trig...using the triple angle formulae to show that...

Hi there

I have this question:

Choosing suitable values of A and B, show that sin + sin θ = 2 sin cos θ

I have been at this one for hours, please could someone give me a nudge in the right direction, maybe show me the first 2 steps that you would take, please do not give me a full solution!

Many thanks
Jackie
Original post by jackie11
Hi there

I have this question:

Choosing suitable values of A and B, show that sin + sin θ = 2 sin cos θ

I have been at this one for hours, please could someone give me a nudge in the right direction, maybe show me the first 2 steps that you would take, please do not give me a full solution!

Many thanks
Jackie


3=2+1

split up the sin3θ\sin{3 \theta} using that.

Then you should be able to get it to look like sin2θcosθ+sinθ(cos2θ+1)\sin{2 \theta} \cos{\theta} + \sin{\theta} ( \cos{2 \theta} + 1)

Next use an identity for cos2θ\cos{2 \theta} and rewrite the 1 in terms of trig functions. You should be able to cancel some tuff and make it look like

sin2θcosθ+2sinθcos2θ\sin{2 \theta} \cos{\theta} + 2\sin{\theta} \cos^2{\theta}

which just needs you to apply one final identity....
(edited 11 years ago)
Reply 2
Yes!!!! I did it...

The rest is:

using 2 sin θ cos θ = sin
so 2 sin θ cos² θ = sin cos θ

sin cos θ + sin cos θ = 2 sin cos θ

thank you so much :smile:
Reply 3
Original post by jackie11
Hi there

I have this question:

Choosing suitable values of A and B, show that sin + sin θ = 2 sin cos θ

I have been at this one for hours, please could someone give me a nudge in the right direction, maybe show me the first 2 steps that you would take, please do not give me a full solution!

Many thanks
Jackie


The wording(Choosing suitable values of A and B) of the question suggest use of the following formula, 2sinAcosBsin(A+B)+sin(AB) 2 \sin A \cos B \equiv \sin (A+B) + \sin (A-B)

So for this question A=2 and B=1.

Original post by latentcorpse
3=2+1

split up the sin3θ\sin{3 \theta} using that.

Then you should be able to get it to look like sin2θcosθ+sinθ(cos2θ+1)\sin{2 \theta} \cos{\theta} + \sin{\theta} ( \cos{2 \theta} + 1)

Next use an identity for cos2θ\cos{2 \theta} and rewrite the 1 in terms of trig functions. You should be able to cancel some tuff and make it look like

sin2θcosθ+2sinθcos2θ\sin{2 \theta} \cos{\theta} + 2\sin{\theta} \cos^2{\theta}

which just needs you to apply one final identity....


Look above.

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