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# Manipulating double angle formula...

Okay I'm trying to do the following question:

The function f is defined by f : x--> 30 sin 3x cos 3x, 0 < x < pi/3.

(a) Write down an expression for f (x) in the form a sin 6x, where a is an integer.

(b) Solve f (x) = 0, giving your answers in terms of p.

Now I spotted you have to use an identity, and it does make sense to use the double angle formula (which the mark scheme also suggests)... however I'm confused as to how to manipulate the double angle formula in order to get the answer for part (a). As the double angle formula is going to be either

sin2x = 2sinxcosx or cos2x = 1- 2sin2x

I'm having difficulty working out where this stands in relation to the fact that everything in the given function is a multiple of 3. Does it then follow that...

sinx3x = 3sinxcosx or that cos3x = 1- 3sin3x ??

... or am I going about the whole thing wrong?

Any help would be greatly appreciated

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Okay I'm trying to do the following question:

The function f is defined by f : x--> 30 sin 3x cos 3x, 0 < x < pi/3.

(a) Write down an expression for f (x) in the form a sin 6x, where a is an integer.

(b) Solve f (x) = 0, giving your answers in terms of p.

Now I spotted you have to use an identity, and it does make sense to use the double angle formula (which the mark scheme also suggests)... however I'm confused as to how to manipulate the double angle formula in order to get the answer for part (a). As the double angle formula is going to be either

sin2x = 2sinxcosx or cos2x = 1- 2sin2x

I'm having difficulty working out where this stands in relation to the fact that everything in the given function is a multiple of 3. Does it then follow that...

sinx3x = 3sinxcosx or that cos3x = 1- 3sin3x ??

... or am I going about the whole thing wrong?

Any help would be greatly appreciated

Well, the first tip is, always look at what the question wants you to do. They say that they want the answer in the form a sin 6x, where a is an integer. Therefore, you would have to use the form sin2x = 2sinxcosx. Now, when you look at 30 sinxcosx, you see that it is already in that form, because it can be written as 15 (2sin3xcos3x). Therefore, the 3x has to be half of the original angle, so that would be 6x. Therefore, it reduces down to 15 sin6x, which is the form that they want. Thus, you see that a = 15. The fact that everything is a multiple of 3 doesn't really matter in this problem. The double angle formula is always sin2x = 2sinxcosx. The angle 2x could be anything. If x = 3x, in this case, then 2x is 6x. If it was 16x, then 2x would be 32 x. In that cause, the double angle formula for sin32x would be 2sin16xcos16.

I hope that all made sense.

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a) 2sin3xcos3x = sin(3x+3x) = sin6x

so 30sin3xcos3x = 15sin6x

b) sin6x=0 when 6x=0 + kpi

x = kpi/6, when k=1, x = pi/6

Edit: whoops, can't think today haha. Thanks Spriteling.

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Irene: sin(pi/2) = 1, so your answer doesn't work.

You'd have to do sin(x) = 0 at pi.

Therefore, when 6x = pi, sin(6x) = 0.

Therefore, x = pi/6.

Thanks you guys