Range of a trig graph?

Hi, this may sound like a ridiculous question because it has given me this:

f(x) = 1-3cos(2x)- 4sin(2x)

And has asked me to find the range. I could differentiate it perhaps and find where the turning points are but it simply asks me to 'state' and it's a 1 mark question so I don't think that's the right way to do it. I thought it was -3 \< f \< 5 but that's wrong. How do I find it? Thanks.

Reply 1

ChanandlerBong

Hi Axon.

You working makes sense but according to my answer sheet the actual answer is -4\< f(x) \< 6 and I'm not sure how they do it. I think your method cannot work in this case because cos(x) and sin(x) can never be 1 at the same value on the x axis because they are a translation of each other?

Reply 2

yusufu

f(x) = 1-3cos(2x)- 4sin(2x) = 1 - {3cos(2x) + 4sin(2x)}
3cos(2x) + 4sin(2x) can be rewritten as Rcos(2x-y)
R = rt(3^2+4^2) = 5
f(x) = 1 - 5cos(2x-y)

should be evident from there.

Reply 3

ChanandlerBong

Oh, I understand it now. I worked out previously that 3cos(x)- 4sin(x) = 5cos(x - 0.9273)

So it's basically, 1 -5 and 1 + 5

-4 \< f(x) \< 6

Reply 4

ChanandlerBong

Thanks for the replies.

Reply 5

gyrase

Gamma

Ah I see, where I went wrong. Somehow I think, this 1-marker question is related to the previous part of the question. If it is a "state that" question, then I would either do a sketch of the function and state the range, or differentiate and find a min and max turning point.

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