Question

Use half angle identity to find the exact value of sin 112.5º?

Answer 1

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Answer 2

Ooo that's colorful ^^

So we can start with this:
\[\Large\rm \sin\left(112.5\right)=\sin\left(\frac{225}{2}\right)\]

Answer 3

But how did you get to sin(225/2)?

Answer 4

So then we need to use our Half-Angle Identity for Sine,\[\Large\rm \sin\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1-\cos(x)}{2}}\]

Answer 5

Yea...I know..I was going to help but since you already startrd...I'm going to let you finish!

Answer 6

Ummmm

Answer 7

I multiplied \(\Large\rm 112.5\) by \(\Large\rm \dfrac{2}{2}\)

The reason I did this is because we really need a 2 denominator to apply our Half-Angle Formula.

Answer 8

\[\Large\rm 112.5\cdot\frac{2}{2}=\frac{112.5}{1}\cdot \frac{2}{2}=\frac{2\cdot112.5}{2}=\frac{225}{2}\]Ok with that part? :o
We got that 2 denominator we were looking for.

Answer 9

Yes, I get that but then you just substitute for your half angle identity?

Answer 10

\[\Large\rm \sin\left(\frac{\color{orangered}{225}}{2}\right)=\pm\sqrt{\frac{1-\cos(\color{orangered}{225})}{2}}\]Yes :)
Is 225 one of our special angles?
Blah I can't remember.
If it's not, we'll have to apply Angle Sum/Difference Formula.

Answer 11

Oh it is, ok good :)

Answer 12

You don't want to leave the plus/minus sign in front though!
Look back at your original angle, and determine which quadrant it is in.
And then try to remember if Sine is positive or negative in that quadrant.

Answer 13

Example:
If we had started with \(\Large\rm \sin(193.4^{\circ})\)
We would only want the `negative` root, because sine is negative in the third quadrant.

Answer 14

The original angle is 112.5, right? which would be in the second quadrant, so it would be positive?

Answer 15

Mmm yes very good. So we only want the positive root.
We can drop the plus/minus sign.\[\Large\rm \sin\left(\frac{\color{orangered}{225}}{2}\right)=\sqrt{\frac{1-\cos(\color{orangered}{225})}{2}}\]

Answer 16

Going back to 225º though, what do you mean by it being a special angle?

Answer 17

Remember that at special places around the unit circle sine and cosine are giving us values like:\[\Large\rm \frac{1}{2}\]\[\Large\rm \frac{\sqrt2}{2}\]\[\Large\rm \frac{\sqrt3}{2}\]Those look familiar? :o

Answer 18

sine of 30, 45 and 60

Answer 19

Mmm yah :)
Those values continue around the circle.

So ummm.. I guess one way you can do it is....\[\Large\rm \cos(225)=-\cos(225-180)=-\cos(45)\]So what I did here is......

Answer 20

you used reference angle

Answer 21

Yah if you remember your reference angles you can do a lot of stuff :p

Answer 22

If this method is too confusing, I apologize >.<
So what I was doing was, spinning halfway around the circle, putting a negative in front since our sign on cosine changed (from Q3 to Q1) and relating it to 45 degrees.

Answer 23

so then cos 45º is sqrt of 3/2?

Answer 24

Nooo you silly billy.
You really gotta try to remember the first quadrant.

45 degrees is pi/4.
I find it easier to remember the angles in radians myself.
All of the pi/4 thingies give us sqrt2/2,
For both sine and cosine.
just have to watch the `sign` on it.

Answer 25

oh yes, its true, i'm sorry!

Answer 26

I think i made a mess though. Can we go back to our half angle identity?

Answer 27

Are we ok up to this point?\[\Large\rm \sin\left(\frac{\color{orangered}{225}}{2}\right)=\sqrt{\frac{1-\cos(\color{orangered}{225})}{2}}\]Or we need to back up a lil further?

Answer 28

thats okay, continue please

Answer 29

support @zepdrix 112.5= 5pi/8

Answer 30

In radians? :)
yah that's way to do it also.

Answer 31

So we're trying to figure out cos(225).
Having trouble there?

Answer 32

no, 225-180, ends up as 45º

Answer 33

so what do we get for cos(225)? :)\[\Large\rm \cos(225)=?\]

Answer 34

-sqrt2/2

Answer 35

Ok great! Let's plug that in.
\[\Large\rm \sin\left(\frac{\color{orangered}{225}}{2}\right)=\sqrt{\frac{1-(-\sqrt2/2)}{2}}\]