Question

v is in the interval (pi/2, pi), sin v=1/15. what is cos(2v)?

Answer 1

Hi gd0x :)

look at your interval. It should be look like this. (pi, pi/2)
or you have missed - sign
(-pi/2, pi)

Answer 2

ok you have
\[\sin(v)=\frac{1}{15}\] and you need
\[\cos(v)\] before you can find
\[\cos(2v)\]

Answer 3

adjacent side is
\[\sqrt{15^2-1^2}=\sqrt{225-1}=\sqrt{224}\]

Answer 4

since you are in quadrant 2, cosine is negative so
\[\cos(v)=-\frac{224}{15}\] and now use
\[\cos(2v)=2\cos^2(v)-1\]

Answer 5

another typo damn
should have been
\[\cos(v)=-\frac{\sqrt{224}}{15}\]

Answer 6

now use
\[\cos(2v)=2\cos^2(v)-1\]

Answer 7

can i give an exact answer using purely numbers?

Answer 8

you should get
\[\cos(2v)=2\times \frac{244}{225}-1\] whatever that is

Answer 9

yes, it is the number i wrote above

Answer 10

\[\frac{223}{225}\]

Answer 11

yep, haha. that was it.

Answer 12

thanks a lot.