please help me solve the trig problem in the image !!!!!!!!!

i know im supposed to use the half angle identity but i know know what to do . i put (1+cos(-3/5)/2 all square roots + and - but im stuck here!

bump bump

>>15339591

The Result ist B.

Greetings from germany

>>15339686

Explaining:
cos(x/2) = +-sqrt(1/2 * (1+cos(x))

But x is in the III Area than x/2 is in the II Area.
And must be negative!


cos(x) = -3/5

cos(x/2) = -sqrt(1/2 * (1 -3/5))
cos(x/2) = -sqrt(1/2 * (2/5))
cos(x/2) = -sqrt(1/5)
cos(x/2) = -sqrt(1/5) = -sqrt(5/25) = -sqrt(5)/5

Oh, I see ..... The correct answer ist D
Sry ^^

Greetings from Germany

File: IMG_20170407_205811089.jpg (542KB, 1440x2560px) Image search: [Google]

542KB, 1440x2560px

hope this helps OP

>>15339722
>But x is in the III Area than x/2 is in the II Area.
>And must be negative!

how do you know x/2 is in quadrant 2 ?

>>15339825
amazing !!!!!!!!!

thank you !

File: 4chan.png (16KB, 793x738px) Image search: [Google]

16KB, 793x738px

>>15339825
Sry, but you're wrong!.
The result must be negative.

>>15339838
yeah listen to this guy OP the bounds were on theta not on the eventual solution cos(thetha/2)