If cosA= -15/17 and A is in quadrant III, find cos 1/2A

- anonymous

If cosA= -15/17 and A is in quadrant III, find cos 1/2A

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- AravindG

ya you have the value of cos A
try to use the formula of cos 2A to get to the answer!!

- AravindG

@sarihmanlapaz did you try that?

- anonymous

what will i use in the 3 formula of cos2a?

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## More answers

- AravindG

just look at the equations and look at what we are trying to get the answer in terms of

- AravindG

we want cos A/2 isnt it?

- AravindG

and we have cos A , now you tell me which equation to use!!

- anonymous

i think cos^2 - 1

- AravindG

\[2\cos^2 x-1\]

- anonymous

so how will i substitute it to that equation?

- hartnn

cosA=2cos^2(A/2)-1
find cos(A/2) from here.

- AravindG

@sarihmanlapaz gt it?

- anonymous

got it! thanks :)

- anonymous

but its none of the choices. >.< how can i solve it?

- hartnn

what did u get ?

- anonymous

0.2435

- hartnn

i m getting -0.2435, is that the option ?

- anonymous

oh no. haha my answer 0.2435 is in the options. i did'nt notice it.

- hartnn

but A/2 will be in 2nd quadrant where cos is negative.
so we put minus
-0.2435

- anonymous

its written in the question that the A is in quadrant 3

- hartnn

yes, A in 3rd,A/2 in 2nd.
is -0.2435 in options?

- anonymous

the options are. A).0.24054 B).0.24125 C) 0.24254 D) 0.24354

- anonymous

pls can you help me?

- hartnn

since all options are positive, i guess A/2 goes in 1st quadrant...then 0.2435 is correct.

- anonymous

no. its wrong my friend. i just check the half angle formula. haha there is a formula for cosA/2 = squareroot of 1/2(1+A)

- hartnn

u mean ....(1+cos A) ?

- anonymous

yes. yes :) i got the right answer 0.2425

- hartnn

how?
the formula we gave and what u wrote are exactly the same...

- anonymous

http://www.themathpage.com/atrig/trigonometric-identities.htm

- anonymous

there is the formula for half angle :)

- hartnn

so using that u got different answer?
we gave u that formula only.....
cosA=2cos^2(A/2)-1
so
cos(A/2)=sqrt{0.5*(1+cos A)}

- anonymous

yeah. thanks i got the right answer.

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