Trig Identities: if tan x = -4/3, pi<x<2pi, find cos(x/2)

Lunare

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May 14, 2008
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Thank you all for your prior help. Once again I need help. Its a tan again.

Suppose tan x = -4/3 with pi < x < 2pi. Find the exact value of cos x/2.

I started this by first finding what quad it terminates by dividing (pi < x < 2pi) / 2 . After that calculating that gives me 90 < (x/2) < 180. So it terminates in Quad II and cos is negative.

after this step I'm lost. I know that tan = sin x / cos x but if i use that i get cos x = sin x /tan x ------> giving me cos x = sin x / (-4/3) and then cos x = -3/4 sin x. but that doesn't look right to me.

I do know that once i get cos x , I can then use the half angle identity of cos (x/2) = +- sqrt (1 - cos x) / 2

I know it might be a simple step I'm missing or something I'm missing completely, but I just don't see it.

Any help would be greatly appreciated.
Lunare



Update.

Can it be tan= sin/cos.
sin is 4 and cos is -3
And then I just plug in -3 into the identity to get cos x/2 ?
 
One way to do it:

since tan(x) is negative and pi<x<2pi - 'x' is in fourth quadrant.

Then how much is sin(x)?

sin(x) = -4/(4^2+3^2)^(1/2) = -4/5

similarly how much is cos(x)

You know

cos(x) = 2*cos^2(x/2) - 1

Find cos(x/2)
 
Lunare said:
Update.

Can it be tan= sin/cos.
sin is 4 and cos is -3<<< No .. NO .. Double NO ...values of sin(x) and cos(x) CANNOT be greater than 1 or less than -1.
And then I just plug in -3 into the identity to get cos x/2 ?
 
royhaas said:
You can use \(\displaystyle tan^2(x)+1 = sec^2(x)\).


Ah I see.
So tan^2 + 1 = sec^2
(-4/3)+ 1 = 1/ cos^2
Cos^2 = 9/25
Cos x = 3/5
Negative cause it terminates in quad ll
Plug it in in the half angle identity
Cos x/2 = - 2/(sqrt 5)


Thanks so much! Now I see it.
 
Subhotosh Khan said:
Lunare said:
Update.

Can it be tan= sin/cos.
sin is 4 and cos is -3<<< No .. NO .. Double NO ...values of sin(x) and cos(x) CANNOT be greater than 1 or less than -1.
And then I just plug in -3 into the identity to get cos x/2 ?
Didn't think so, but a math major said i could. Thank you!
 
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