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# Trig Identity Watch

1. Prove :
1)
1 - sin theta over cos theta SAME AS cos theta over 1 + sin theta

2) 1 - 2sin^2 theta over cos theta + sin theta same as cos theta - sin theta .
--------------

I'm stuck on this because for one I can't make the 1-sin theta into cos as its not squared so I'm stuck on what to do .

For two, Im not sure what 1 - 2sin^2 theta is, I can do 1- sin^2 theta as its cossquared, but not sure what to do with the 2
2. Just reaarange:

(1-sinx)/cosx = cosx/(1+sinx) (i would have used equivalent sign but there isn't one)

=> (1+sinx)(1-sinx) = (cosx)^2
=> 1 - (sinx)^2 = (cosx)^2

Which using sin^2 + cos^2 = 1 shows it's true.
3. (Original post by nicatre)
Just reaarange:

(1-sinx)/cosx = cosx/(1+sinx) (i would have used equivalent sign but there isn't one)

=> (1+sinx)(1-sinx) = (cosx)^2
=> 1 - (sinx)^2 = (cosx)^2

Which using sin^2 + cos^2 = 1 shows it's true.
Is that cross multiplying ? Not sure how its meant to be rearranged ?

Thanks =)
4. (Original post by Gelato)
Is that cross multiplying ? Not sure how its meant to be rearranged ?

Thanks =)
Yeah, it's just rearranging equal fractions

Like if a/b = c/d then ad = bc
5. (Original post by nicatre)
Yeah, it's just rearranging equal fractions

Like if a/b = c/d then ad = bc
I got stuck on it on the part now that

I end up with:

1-sin^2 theta = cos^2theta SAME AS cos theta over 1+sintheta

Not sure what to do now, I can move the - sin squared theta to get 1 on the left hand side, but not sure how it equals the other side ?
6. (Original post by Gelato)
I got stuck on it on the part now that

I end up with:

1-sin^2 theta = cos^2theta SAME AS cos theta over 1+sintheta

Not sure what to do now, I can move the - sin squared theta to get 1 on the left hand side, but not sure how it equals the other side ?
u have 1 - sin^2 = cos^2?

If you have that, then that proves it.

You can plug in a value of theta if needed just to verify it.
7. (Original post by nicatre)
u have 1 - sin^2 = cos^2?

If you have that, then that proves it.

You can plug in a value of theta if needed just to verify it.
Thanks alot, i somehow made the beginning one come back in my working out, all fine now. Can you help on the 2nd aswell please ?
8. (Original post by Gelato)
Thanks alot, i somehow made the beginning one come back in my working out, all fine now. Can you help on the 2nd aswell please ?
What is the second one?
9. (Original post by nicatre)
Just reaarange:

(1-sinx)/cosx = cosx/(1+sinx) (i would have used equivalent sign but there isn't one)

=> (1+sinx)(1-sinx) = (cosx)^2
=> 1 - (sinx)^2 = (cosx)^2

Which using sin^2 + cos^2 = 1 shows it's true.
Woah woah, you shouldn't be doing this to prove a Trig Identity. If it's an identity you stick with one side, either the LHS or RHS, and manipulate it to get it to equal the other side. You most certainly shouldn't be crossing over between the two sides, adding and multiplying things.

So you have to show

You need to employ a similar idea when you rationalise denominators with surd in. Try multiplying the right hand side by
Note that this is basically equal to 1, so you're not changing your identity in any way. This should then be fairly easy to manipulate into the LHS, as you will get sin squared and you can then use your other identities.
10. listen to dnkt.

If you're doing C3 trig identities, you have to make the LHS equal the RHS through manipulation, or RHS to LHS.
11. (Original post by ?!master?!mini?!)
listen to dnkt.

If you're doing C3 trig identities, you have to make the LHS equal the RHS through manipulation, or RHS to LHS.
Actually you don't. Although it is certainly considered best practice to work on one side, full marks will still be awarded if both sides are manipulated.
12. (Original post by Mr M)
Actually you don't. Although it is certainly considered best practice to work on one side, full marks will still be awarded if both sides are manipulated.
I see.

I was only taught to manipulate either the lhs or rhs.

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