Question

Find the exact value for sin((3pi/4) + (5pi/6))

Answer 1

use sum identity for sin

Answer 2

that is
sin(x+y)=sin(x)cos(y)+sin(y)cos(x)

Answer 3

then you would use the unit circle to evaluate
sin(3pi/4)
sin(5pi/6)
cos(3pi/4)
cos(5pi/6)

Answer 4

plug in
and follow the order of operations

Answer 5

When using the unit circle to find your values, find your reference angles.

Answer 6

So would that be like this - > sin(3pi/4 + 5pi/6) = sin(3pi/4)cos( 5pi/6) + sin(5pi/6)cos(3pi/4) //Sorry for the slow reply

Answer 7

oh yes

Answer 8

sorry I have multiple windows open :p

Answer 9

It's fine, and then like you said I'd just follow order of operations?

Answer 10

yep once you used the unit circle to evaluate all of those thingys

Answer 11

Um, well I feel dumb, but how do you do order of operations with sin/cos? Like do I just solve for the sin / cos values and then add them and compare? Or is it more involved than that?
Also, Jhanny how do the reference angles figure into find the exact values? //I've been kinda lacking in my Trig studies.

Answer 12

@mutedwolf you have to used the unit circle to evaluate
sin(3pi/4)
sin(5pi/6)
cos(3pi/4)
cos(5pi/6)
then use order of operations

Answer 13

for example
sin(3pi/4)=sqrt(2)/2

Answer 14

Ohhh, okay, that makes more sense.

Answer 15

Yeah....that's why I drew you a unit circle, lol.

Answer 16

Ah, lol, okay. So would the evaluated equation look like this? ->
(sqrt (2)/2 + sqrt 1/2) = (sqrt (2)/2 * -sqrt 3/2) + sqrt 1/2 * -sqrt 2/2)

Answer 17

to the left what you have is off
but to the right it looks good

Answer 18

You were asking about the reference angles, totally skipped over that post. Sorry. The reference angles tell you the x an y values in the other 3 quadrants. Because the reference angles are all positive, it's easier to use them than calculating what the x and y components in the other quadrants will be - just look at quadrant one for all your answers :)

Answer 19

The only thing that changes is their sign

Answer 20

\[\sin(\frac{3\pi}{4}+\frac{5\pi}{6})=\frac{\sqrt{2}}{2} \frac{-\sqrt{3}}{2}+\frac{1}{2} \frac{-\sqrt{2}}{2}\]
just wanted to change what you had above
because it was icky to me :p

Answer 21

Ohh, I was getting mixed up about left geez, thought you meant one of my unit circle values was off.

Answer 22

no I just want you to know sin(x+y) doesn't equal sin(x)+sin(y)
and that is what it look like what you were doing on the left

Answer 23

\[\text{ so you actually found what } \sin(\frac{3\pi}{4}+\frac{5\pi}{6}) \text{ or } \sin(\frac{19}{12} \pi) \text{ equals exactly }\]

Answer 24

I would multiply a little and combine the fractions though

Answer 25

Thanks for the help!