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Welcome to NRICH.

sinq-cosq

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By Anonymous on Tuesday, January 23, 2001 - 01:05 am:

Can somebody please tell me how to graph sinq-cosq

I really don't have a clue how to do it. All the help greatly appreciated. Thanks.

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By Tim Martin (Tam31) on Tuesday, January 23, 2001 - 01:27 am:

The trick is to rearrange into the form of a single trig expression. It is possible to write anything of the form Asinq + Bcosq as Rcos(q+a)

R cos(q + a) = sinq - cosq

Expanding using the identity cos(A+B)=cosAcosB - sinAsinB:

Rcosqcosa - Rsinqsina = sinq - cosq

Therefore

Rsina = -1
Rcosa = -1

Þtana = 1
Þa = p/4

ÞRsin(p/4) = -1
ÞR = -2^1/2

Putting this together:

sinq - cosq = -2^1/2cos(q + p/4)

Hope that makes some sort of sense, ask if you aren't sure about the reasoning.

Tim

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By Anonymous on Tuesday, January 23, 2001 - 07:02 pm:

Thanks, it makes sense now!!!

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