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# Trig question :( Watch

1. Show that 2Sin2X - 3Cos2X - 3SinX + 3 = SinX(4CosX+6SinX - 3)

hmm.. Where do I start?

I'd appreciate any help

Thanks.
2. (Original post by Jonario)
Show that 2Sin2X - 3Cos2X - 3SinX + 3 = SinX(4CosX+6SinX - 3)

hmm.. Where do I start?

I'd appreciate any help

Thanks.
Multiply out the RHS and note that and .
3. (Original post by Jonario)
Show that 2Sin2X - 3Cos2X - 3SinX + 3 = SinX(4CosX+6SinX - 3)

hmm.. Where do I start?

I'd appreciate any help

Thanks.
Presumably you know the double angles identities for sin 2x and cos 2x?
4. (Original post by Farhan.Hanif93)
Multiply out the RHS and note that and .
I got 4cosXSinX+6sin^2X - 3sinX

but you see when you say note.. Does that mean I have to rearrange the LHS?
5. (Original post by Mr M)
Presumably you know the double angles identities for sin 2x and cos 2x?
Yes I do

2 Sin 2X = 2Sin2XCos2X?
6. (Original post by Jonario)
Yes I do

2 Sin 2X = 2Sin2XCos2X?
No need for that first 2 and the 2X is just X on the RHS.
7. (Original post by Jonario)
Yes I do

2 Sin 2X = 2Sin2XCos2X?
If this were true cos(2x)=1.....always.
8. (Original post by Yellowmonkeyman)
No need for that first 2 and the 2X is just X on the RHS.
Oh man i'm so confused now.

Do I start with the RHS or the LHS, do i even change the LHS?
9. (Original post by Jonario)
Oh man i'm so confused now.

Do I start with the RHS or the LHS, do i even change the LHS?
No. What you do here is take ONE side of the equation and make it look the same as the other without changing the other side itself. You need to do this using identities.
10. (Original post by Farhan.Hanif93)
No. What you do here is take ONE side of the equation and make it look the same as the other without changing the other side itself. You need to do this using identities.
Ok so i get the RHS SinX(4cosX+6SinX-3) and multiply out.

I get 4cosXSinX+6Sin^2X - 3Sinx

Then I use the double angle formulae .. but I can't point out which one I use
11. (Original post by Farhan.Hanif93)
No. What you do here is take ONE side of the equation and make it look the same as the other without changing the other side itself. You need to do this using identities.
The LHS has 2sin2X so I use the SIN2A formulea? but how do I convert the RHS to this?
12. (Original post by Jonario)
Ok so i get the RHS SinX(4cosX+6SinX-3) and multiply out.

I get 4cosXSinX+6Sin^2X - 3Sinx

Then I use the double angle formulae .. but I can't point out which one I use
Well I've told you that . If you notice that . What is that equal to?
Also do you agree that is equal to what you got when you expanded out the RHS?
Then perhaps notice that .
13. (Original post by Farhan.Hanif93)
Well I've told you that . If you notice that . What is that equal to?
Also do you agree that is equal to what you got when you expanded out the RHS?
Then perhaps notice that .

How Is that true?
14. (Original post by Jonario)

How Is that true?
Of course that's true, that's simple multiplication. No trig identities involved at all. if you have 4 apples, thats the same as having 2 lots of two apples.
15. (Original post by Jonario)

How Is that true?
err...because 2*2=4 and sinx*cosx=cosx*sinx or did you mean something else?

(note that I've used the star* as the sign for multiplication)
16. 2sin2x = 4sinxcosx
cos2x= 1- sin^2 x

now expand:
4sinxcosx-3+6sin^2 x -3sinx +3
.... and factorise

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