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# c3, simple trig question watch

1. how can you get sin2x - tanx to

2cos2x?
2. sin2x - tanx
=2sinxcosx - sinx/cosx
=sinx(2cosx - 1/cosx)
=sinx[(2cos²x-1)/cosx]
=sinx[(cos2x)/cosx]
=tanxcos2x

hmmmm maybe not that way :s
3. i got 2 tanXcos2X...dnt knw how toget furtha??
4. Q must b wrong
5. (Original post by nas7232)
how can you get sin2x - tanx to

2cos2x?
What is the full question? to prove that they are equal?

If so: sin2x - tanx

=> 2sinxcosx - sinx/cosx

=> [2sinxcos2x - sinx]/cosx

=> [sinx(2cos2x - 1)]/cosx

=> tanxcos2x

Now tanx =/= cos2x so I guess I have done something wrong or the question is wrong.
6. (Original post by mackin boi)
i got 2 tanXcos2X...dnt knw how toget furtha??
hmmm how u get the 2? Can u show ur working?
7. i cant see how its possible either
8. (Original post by samdavyson)
What is the full question? to prove that they are equal?

If so: sin2x - tanx

=> 2sinxcosx - sinx/cosx

=> [2sinxcos2x - sinx]/cosx

=> [sinx(2cos2x - 1)]/cosx

=> tanxcos2x

Now tanx =/= cos2x so I guess I have done something wrong or the question is wrong.
ops sorry, that was part a) done right

part b is sin2x - tanx = 2cos2x
solve that equation
in the interval of 0<x<180

Sorry! i typed the question out wrong lol
9. (Original post by Syncman)
hmmm how u get the 2? Can u show ur working?
The question must be a mistake.

I don't think there should be a 2 in that either.
10. ah i can do it now...what a waste of thread
11. (Original post by nas7232)
how can you get sin2x - tanx to

2cos2x?
sin2x - tanx
= 2sincosx - sinx/cosx
= (2sinxcos^2x - sinx)/cosx
= {2sinx[(cos2x + 1)/2] - sinx}/cosx
= {sinx(cos2x + 1) - sinx}/cosx
= {sinxcos2x + sinx - sinx}/cosx
= sinxcos2x/cosx
= tanx.cos2x

12. ops sorry, that was part a) done right

part b is sin2x - tanx = 2cos2x
solve that equation
in the interval of 0<x<180

Sorry! i typed the question out wrong lol
.
13. (Original post by nas7232)
ah i can do it now...what a waste of thread
I still can't see how to solve it.

(Original post by nas7232)
ops sorry, that was part a) done right

part b is sin2x - tanx = 2cos2x
solve that equation
in the interval of 0<x<180

Sorry! i typed the question out wrong lol
Ahh. Thank the heavens.

sin2x - tanx = 2cos2x

=> cos2xtanx = 2cos2x

=> cos2x = 0

OR => tanx = cos2x

=> sinx/cosx = cos2x - sin2x

=> Then what?
14. part b is sin2x - tanx = 2cos2x
solve that equation
in the interval of 0<x<180

as you worked out:

sin2x - tanx = tanx.cos2x

tanx.cos2x = 2cos2x
tan.cos2x - 2cos2x = 0
cos2x(tanx - 2)=0
cos2x = 0
or tanx = 2

solutions
45,63.4,135

once again, sry for not typing it out properly. Now i know why i couldn't initially do it.

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