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 ?
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 ?