Question

If cosx=2/3 and x is in quadrant 4, find
sin(x/2)

Answer 1

What is your plan for a solution?

If x is in quadrant 4 \(270º < x < 360º\), where does that put x/2?

Answer 2

not sure? halfway between 270 and 360?

Answer 3

315?

Answer 4

Divide all three elements.

\(270º < x < 360º\)

\(135º < x/2 < 180º\)

It is in the last half of the Quadrant II. Make sense?

Answer 5

okay I am following so far

Answer 6

What is the nature of sine and cosine in Quadrant II?

Cosine is negative and sine is positive. Don't forget this! We'll get it wrong if we do. Fair enough?

Answer 7

Yes got it!

Answer 8

Now, we need a half angle formula. Have you one of those?

Answer 9

Waiting... hoping... looking it up... \(\sin(x/2) = \pm\)... Are we close?

Answer 10

well I only have double angle identities in my notes
which are
Cos2x=1-2sin^2x, cos^2x-sin^2x, 2cos^2x-1
sin2x=2sincosx
tan2x=(2tanx)/1+tan2x

Answer 11

but I am not that sure about half identitites because I dont see them anywhere in my notes, just squared, double, sum, and difference identitities

Answer 12

That's good enough. Use one of those cosine ones. We need sin(x/2), so do a little algebra.

\(\cos(2x) = 1 - 2\sin^{2}(x)\)

Use a different angle... y = 2x to give...

\(\cos(y) = 1 - 2\sin^{2}(y/2)\) -- We just invented a half-angle formula!! Still with me?

Answer 13

Okay got it. I didnt know you could do that. But why the (y/2) I dont see how that came from plugging in the y=2x to the initial double angle identity for cos?

Answer 14

because there is only an x there?

Answer 15

y = 2x ==> y/2 = x

Just division by 2.

Answer 16

Oh I see! makes sense

Answer 17

x, y, q, Fred -- It is of no consequence. It's just a label.

Let's solve for ... \(\sin(y/2) = \pm\sqrt{\dfrac{1\cos(y)}{2}}\)

Just algebra. Nothing tricky.

Answer 18

thanks for going through this in detailed steps btw, this is exactly what I needed! this stuf is hard.

Answer 19

Hmm so square both sides
then multiply both sides by 2
Then divide by one and
cos(y)=(sin^2 2(x))/1

Answer 20

Tough question. What does that "\(\pm\)" symbol mean?

Answer 21

Well in this case we either have to chose the negative or possitive answer based on the quadrant the answer is in

Answer 22

That silly symbol means about six different things in all of mathematics.

Excellent!! In this case, that \(\pm\) mean to pick the correct sign. In this case, we are in Quadrant II and we are talking about the sine, so pick \(+\).

Substitute the known value of cosine and do the arithmetic.

Answer 23

and in this case we found out it is in the second quadrant and we are focussing on sin so it is only the possitive

Answer 24

Yes:)

Answer 25

SO it would be cosy=sin^2 2x/1

Answer 26

Whoops!!! I didn't type it correctly. \(\sin(x/2) = \pm\sqrt{\dfrac{1 - \cos(x)}{2}}\).

Sorry about that.

Answer 27

Oh so we are using x instead?

Answer 28

but then wouldnt it be the same but with x
o cos(x)=(sin^2 2x)/1

Answer 29

Doesn't matter. The original problem statement gave us 'x', so we may as well go with it. Don't try to transform back to the double angle. Think of it as

\(\sin(HalfofSomeAngle) = \sqrt{\dfrac{1 - \cos(SomeAngle)}{2}}\)

It does not matter what you call "some angle". Just go with it. Whether it is x and x/2 or z/5 and z/10 or 12w and 6w - as long as the one outside the radical is half the size of the one inside the radical. It's the same result.

\(\cos(x) = 2/3\) so \(\sin(x/2) = \sqrt{\dfrac{1 - 2/3}{2}}\)

I'll leave you the arithmetic. gtg