Quick Trig qu (1 Viewer)

[currysauce]

Solve

sinx = tan (x/2) for 0 <= x <= 2pi

thanks

 

[acmilan]

t-method:
t = tan(x/2)
sin(x) = 2t/(1+t²) (if i remember right?)

So 2t/(1+t²) = t

Solve that for t, then make tan(x/2) = solution and you can find x

 

[currysauce]

u know whats weird, is this came from a 2unit paper... a trial

so i didn't think of the t method

hmm... thanks

 

[FinalFantasy]

t-method:
t = tan(x/2)
sin(x) = 2t/(1-t²) (if i remember right?)

So 2t/(1-t²) = t

Solve that for t, then make tan(x/2) = solution and you can find x

isn't it 2t\(1+t²) for sin?

 

[currysauce]

yeah, but that doesn't help me find a valid reason for this in a 2unit paper

 

[FinalFantasy]

oh the t- formula isn't in 2unit?

 

[acmilan]

FinalFantasy: youre most likely right, i remembered it was in that form. Ive never had to, and probably never will, use the t-method.

 

[FinalFantasy]

Solve

sinx = tan (x/2) for 0 <= x <= 2pi

sinx=sin(x\2)\cos(x\2)
2sin(x\2)cos(x\2)=sin(x\2)\cos(x\2)
2sin(x\2)cos²(x\2)=sin(x\2)
2sin(x\2)cos²(x\2)-sin(x\2)=0
sin(x\2)(2cos²(x\2)-1)=0
.: (x\2)=0 or cos²(x\2)=1\2---->cos(x\2)=+-1\sqrt(2)

im sure u can do that in 2unit?

 

[currysauce]

hmm that looks good, thanks man

sinx=sin(x\2)\cos(x\2)
2sin(x\2)cos(x\2)=sin(x\2)\cos(x\2)

wtf, explain that please

 

[FinalFantasy]

sin2x=2sinxcosx
that's analogous to what i did

 

[currysauce]

well thats clearly not 2unit, double angles

 

[acmilan]

haha you still used 3 unit methods

ie. the double angle formula

 

[acmilan]

One way you could do it, which may be sufficient for 2 unit, is to plot y = sinx and y = tan(x/2) and notice that the only place they meet is at 0, pi and 2pi. Not very elegant though

 

[FinalFantasy]

hey but there was a 2unit exam some time last term i think and this sort of thing was in it..