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expscv
Apr11-04, 02:04 AM
how do i find intersection of sin(x) and cos(x)? wat method do i use?

cookiemonster
Apr11-04, 02:06 AM
sinx = cosx
sinx/cosx = 1
tanx = 1
x = arctan(1)
x = pi/4

cookiemonster

himanshu121
Apr11-04, 02:22 AM
how do i find intersection of sin(x) and cos(x)? wat method do i use?

Apart from it u can do it graphically. But Still u have to do wat cookie monster( :redface: ) has done

expscv
Apr11-04, 02:33 AM
but wat if is sin(x) and cos(2x) ?

cookiemonster
Apr11-04, 02:40 AM
That's a little more difficult. You'd have to use a half-angle formula and solve it similarly.

cookiemonster

expscv
Apr11-04, 03:04 AM
with ur help it seems to be

sin(x)= 1- 2sin(x)^2

2sin(x)^2+sin(x)-1=0

hey it works thx all

expscv
Apr11-04, 04:42 AM
wait but how do i solve 2sin(x)^2=tan(x)

Chen
Apr11-04, 04:48 AM
Use the identity:
\tan ^2 x + 1 = \frac{1}{\sin ^2 x}

expscv
Apr11-04, 04:56 AM
omg i dont reallyget how this identity could help me~

cookiemonster
Apr11-04, 05:04 AM
Eliminate the sin^2(x) with that identity.

cookiemonster

Chen
Apr11-04, 05:09 AM
That identity should give you:

4\sin ^6 x + \sin ^2 x - 1 = 0

Now let t = sin2x and solve the equation.

(I eliminated tanx rather than sinx.)

expscv
Apr11-04, 05:10 AM
yeah but it trun out to be tan(x)^3+tan(x)-2=0

cookiemonster
Apr11-04, 05:12 AM
So now you gotta do some more factoring. More fun algebra!

Edit: Fine!

cookiemonster

Chen
Apr11-04, 05:14 AM
I think it's -2...

expscv
Apr11-04, 05:20 AM
god i m having a headache with everything

expscv
Apr11-04, 05:23 AM
thx all , i do this after i wake up tommor

expscv
Apr11-04, 05:28 AM
wait tan^2+1= 1/cos^2 is it?

Chen
Apr11-04, 05:31 AM
No, \tan ^2 x + 1 = \frac{1}{\sin ^2 x}.

matt grime
Apr11-04, 05:45 AM
Chen, you might want to check that, as tan of 0 is not infinity.

Chen
Apr11-04, 05:54 AM
Of course you are right.

\tan ^2 x + 1 = \frac{1}{\cos ^2 x}