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# Help please - expressing as Rcos(2x+a) watch

1. Afternoon all,

I'm using the book Advanced Maths for AQA, Core Maths C3/C4 to study from home. I'm stuck on a question though and it doesn't really give me any prev examples to show how it is done: -

9E 5a) Express 9cos2x-4sin2x in the form Rcos(2x+a)

I've been doing loads previously but none of them had a double angle in. Please can somebody help?

The answer in the book is root(97)cos(2x+24) the 24 being in degrees.

Many Thanks,
Ross

I've finished studying for C3 and now I have C4 and MFP3 to study before my A2 exams in June, I'll have done an A level in 10 months if I manage to crack it in June (started in Aug 10). Hadn't studied maths since school a few years ago but realised half way through my NC in mech engineering that I need the A level to get in Uni and I need grade B to meet my offer.

Got a B at AS, just!
2. Let . Now it's not a double angle
3. (Original post by rosschambers1987)
Afternoon all,

I'm using the book Advanced Maths for AQA, Core Maths C3/C4 to study from home. I'm stuck on a question though and it doesn't really give me any prev examples to show how it is done: -

9E 5a) Express 9cos2x-4sin2x in the form Rcos(2x+a)

I've been doing loads previously but none of them had a double angle in. Please can somebody help?

The answer in the book is root(97)cos(2x+24) the 24 being in degrees.

Many Thanks,
Ross

I've finished studying for C3 and now I have C4 and MFP3 to study before my A2 exams in June, I'll have done an A level in 10 months if I manage to crack it in June (started in Aug 10). Hadn't studied maths since school a few years ago but realised half way through my NC in mech engineering that I need the A level to get in Uni and I need grade B to meet my offer.

Got a B at AS, just!

You know that the rate of cosa and sin a is 9:4, because taking out any factor
from the two terms this rate will be constant.
You should to take out factor because cosa would not be equal with 9 or sina with 4.
Let

Taking square both equation and adding them together you will get value of t,
and from that 9t as cosa or 4t as sina. For this multiply by 1/t factor the terms.
4. 9cos2x-4sin2x in the form Rcos(2x+a)

9cos2x-4sin2x=Rcos2xcosa-Rsin2xsina

Rcosa=9....(1)
Rsina=4.....(2)

Rsina/Rcosa=4/9

So
tana=4/9 (the R's cancel out)

a=23.96...
a~24
R=square root((9)^2+(-4)^2)
=square root 97

Therefore --->>> square root (97)cos(2x+24)
5. Hi guys,

Thanks very much, I had been doing it for hours so maybe my brain wasn't at its optimum but now you have explained it's easy

Thanks again, I'm about to post another question, so if you fancy helping me out there then I'd be very grateful!

The question I was going to ask is how to integrate sin^3(2x)

I tried using the substitution method but can't seem to get it to work....
6. I'm not sure if this is right but I gave it a go...

Integral of sin^3(2x) dx

well first you can factorise sin(2x) to get a positive power --->> the integral of sin(x)[(sin^2(2x)]

then use the trig. identity sin^2(x)+cos^2(x)=1 so the integral of (1-cos^2(2x))sin(x) dx

now use substitution, u=cos(2x) the derivative of that is -2sin(2x)

there dx=du/-2sin(2x)

so the integral of (1-u^2)sin(2x) du/-2sin(2x)

you can cancel out the sin(2x) function and factorise out the negative half to get

-1/2 integral 1-u^2 du

then just integrate to get -1/2(u-u^3/3)+c

then you get -1/2cos(2x)- 1/3cos^3(2x)+c

I may be wrong so check out this video on youtube and see if you can follow the steps

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