Question

if cosx 5/13 and sinx < 0, find sin(x/2)

Answer 1

sorry, its cosx = 5/13

Answer 2

Shall we work together on this?

Answer 3

sure, why not? :D

Answer 4

Awesome! Let's do this. Let me warn you: this will take a little bit of time to do.
http://www.twiddla.com/584559

Let's go here because trig is better done on "paper".

Answer 5

This is a 5 12 13 Pythag triangle and Sin is negative in 3 and 4 quadrant....

Answer 6

my book's answer is actually negative... which is why im asking this question

Answer 7

Wait, negative or positive?

Answer 8

Should be negative, as it's in the 4th quadrant

Answer 9

You could use the identity \[cos(x)=1-2sin^2\left(\frac{x}{2}\right)\]

Answer 10

Or draw the appropriate triangle and just bisect the angle..

Answer 11

this is what you do if you want easy way. look up "half angle formula" for sine, which says
\[\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}\]

Answer 12

replace
\[\cos(x)\] by
\[\frac{5}{13}\] because that is what you are told. you are also told that you want the negative one. substitute and be done

Answer 13

get
\[-\sqrt{\frac{1-\frac{5}{13}}{2}}\] then some arithmetic to get
\[-\sqrt{\frac{\frac{8}{13}}{2}}=-\sqrt{\frac{4}{13}}=-\frac{2}{\sqrt{13}}\]\]

Answer 14

@Satellite: We got that too, but his/her book says it's the same thing in positive.
So the book is wrong?

Answer 15

oh now i have to think. hold on while i turn on my battery

Answer 16

ok it helps if i read. it says
\[\sin(x)<0\] not
\[\sin(\frac{x}{2})<0\]

Answer 17

cosine is positive, sine is negative. you are in quadrant IV (don't you love roman numerals?)

Answer 18

so when you take half of x you will be somewhere up in quadrant II and sine will be positive. so it is positive. sorry

Answer 19

answer is the positive root not negative one. my fault.

Answer 20

i dont really understand the signs...

Answer 21

it starts at quadrant IV, so shouldn't the formula be\[\sin (x/2) = -(\sqrt{1-cosx}/2)\]?