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Further Trigonometric identities C3 HELP!

using the formula:

Rsin(x−α)=Rsinxcosα−RsinαcosxRsin(x- \alpha )= Rsinxcos\alpha - Rsin \alpha cosx

I got

Rsinα=3Rsin\alpha= \sqrt 3 , Rcosα=1Rcos\alpha= 1

tanα=3tan\alpha=\sqrt 3

α=π2 \alpha = \frac{\pi}{2}

then I did some simple algebra to get:

R=3sinπ2 R= \frac{\sqrt 3}{sin\frac{\pi}{2}}
OR
R=1cosπ2 R= \frac{1}{cos\frac{\pi}{2}}


To get: R=2 R=2


But the books shows a completely different method by which it squares the R in the original equation, is what I'm doing wrong and more importantly will I lose marks for it in the exam:rolleyes:

Thanks in advance:smile:
Reply 1
Avatar for CMM
CMM
4
Square-root of 3 over sin Pi/2?

= 2??

What?

I don't understand how you came to that either..

I do find it easier to square and add both.

(1)^2 +(sqrt3)^3 = R^2(sin^2a + cos^2a)

4 = (1)R^2

--> R = 2
Reply 2
Avatar for CMM
CMM
4
Oh I see, typo. Pi/2 you mean a?

Seems to work well enough, perhaps it's taught the other way so people don't mess up going from tan(a) to sin(a) ?
In2deep
using the formula:

Rsin(x−α)=Rsinxcosα−RsinαcosxRsin(x- \alpha )= Rsinxcos\alpha - Rsin \alpha cosx

I got

Rsinα=3Rsin\alpha= \sqrt 3 , Rcosα=1Rcos\alpha= 1

tanα=3tan\alpha=\sqrt 3

α=π2 \alpha = \frac{\pi}{2}

then I did some simple algebra to get:

R=3sinπ2 R= \frac{\sqrt 3}{sin\frac{\pi}{2}}

R=2 R=2


But the books shows a completely different method by which it squares the R in the original equation, is what I'm doing wrong and more importantly will I lose marks for it in the exam:rolleyes:

Thanks in advance:smile:

To get R, I've been shown the other way of finding R.

You have: RSinα=3RSin{\alpha} = \sqrt{3} and RCosα=1RCos{\alpha} = 1
So then you get R2Sin2α=3R^2Sin^2{\alpha} = 3 and R2Cos2α=1R^2Cos^2{\alpha} = 1
So: R2Sin2α+R2Cos2α=R2(Sin2α+Cos2α)=4R^2Sin^2{\alpha} + R^2Cos^2{\alpha} = R^2(Sin^2{\alpha} + Cos^2{\alpha}) = 4

And what do you know about Sin2α+Cos2αSin^2{\alpha} + Cos^2{\alpha}?
Reply 4
Avatar for In2deep
In2deep
OP
14
Wednesday Bass
To get R, I've been shown the other way of finding R.

You have: RSinα=3RSin{\alpha} = \sqrt{3} and RCosα=1RCos{\alpha} = 1
So then you get R2Sin2α=3R^2Sin^2{\alpha} = 3 and R2Cos2α=1R^2Cos^2{\alpha} = 1
So: R2Sin2α+R2Cos2α=R2(Sin2α+Cos2α)=4R^2Sin^2{\alpha} + R^2Cos^2{\alpha} = R^2(Sin^2{\alpha} + Cos^2{\alpha}) = 4

And what do you know about Sin2α+Cos2αSin^2{\alpha} + Cos^2{\alpha}?


Thanks:smile: I get that and it produces the same result as I got with my method. Was my method a fluke and will it not always work? is this the only right method?
In2deep
Thanks:smile: I get that and it produces the same result as I got with my method. Was my method a fluke and will it not always work? is this the only right method?

I think your method is a valid other method. But I haven't been taught that. I just know how to prove the way I've done.
In2deep
Thanks:smile: I get that and it produces the same result as I got with my method. Was my method a fluke and will it not always work? is this the only right method?

I think I see what you've done, you've put the alpha you've just found into coefficient equations and then made R the subject. This is fine but the drawback would be that if you made a mistake in finding alpha then used that value to find R then you will lose a lot of marks. It's safer to square and add.
Reply 7
Avatar for In2deep
In2deep
OP
14
Farhan.Hanif93
I think I see what you've done, you've put the alpha you've just found into coefficient equations and then made R the subject. This is fine but the drawback would be that if you made a mistake in finding alpha then used that value to find R then you will lose a lot of marks. It's safer to square and add.


The answer I was looking for:smile: I DON'T want to lose any marks so I'll play it safe.

Thanks everyone:top:

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