a simple 3u q's (1 Viewer)

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sorry, i posted here cos a lot of ppl dun reply in 3u

well, i got a simple trig q's..

show that if 0<x< pi/4,
then sin2x > 2sin^2 x

i hate trig

 

[Rahul]

i'll give it a shot...

sin2x > 2sin^2x

using sinx=2sinxcosx,

2sinxcosx=2sin^2x
cosx=sinx
sinx=cosx

divide thru by cosx,

tanx=1
x=tan^-11
x=45 or pi/4

edit: whats the point of 0 < pi/4?

 

[ND]

sin2x - (sinx)^2 = 2sinxcosx - (sinx)^2
= sinx(2cosx-sinx)
sinx > 0 for 0 < x < pi/4 because cosx>sinx for 0 < x < pi/4, 2cosx-sinx>0.
.'. sin2x - (sinx)^2 > 0
sin2x > (sinx)^2.

edit: hehe even not 4u students have resorted to posting 3u q's here.

 

[ND]

Originally posted by Rahul

2sinxcosx=2sin^2x
cosx=sinx

Just a note: when solving these, you shouldn't divide through like that, you lose solutions.

 

[McLake]

Originally posted by ND
Just a note: when solving these, you shouldn't divide through like that, you lose solutions.

If you do devide through, check sinx=0 as a solution ...

 

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ahh..thx a lot...

 

[Rahul]

Originally posted by ND
Just a note: when solving these, you shouldn't divide through like that, you lose solutions.

Originally posted by McLake
If you do devide through, check sinx=0 as a solution ...

thanks