Question

Finding the sinΘ?! Please!

Answer 1

YES!!!

Answer 2

lol, doesn't anyone have a unit circle cheat sheet?

Answer 3

O know x = -2 and r = -5

Answer 4

*I know

Answer 5

use pythagoras to find \(b\)
then it is \(\sin(\theta)=\frac{b}{5}\)

Answer 6

This is my introduction to Trig, and I'm having a hard time following. My lesson says nothing about "b" :(

Answer 7

They also have nothing on triangles..

Answer 8

you want the other side of the triangle,
ok we can do it the unit circle way too if you like, but first lets to it the triangle way

Answer 9

okay, thank you

Answer 10

pythagoras gives you
\[2^2+b^2=5^2\] so
\[b^2=5^2-2^2\] or
\[b=\sqrt{25-4}=\sqrt{21}\]

Answer 11

then by good old right triangle trig, is it "opposite over hypotenuse" i.e.
\[\sin(\theta)=\frac{\sqrt{21}}{5}\]
not that it is positive because tangent and cosine are both negative, so the sine must be positive

Answer 12

*note

Answer 13

oh damn, sorry
it says tangent is POSITIVE that means the answer has to be
\[-\frac{\sqrt{21}}{5}\]

Answer 14

ooooohh my gosh. Okay that's where \[\sqrt{21}\] was coming from. So will I be needing to review geometry to continue with trig? that's not too bad!

Answer 15

want to do it unit circle way? we get the same answer, just more annoying arithmetic

Answer 16

Yes, actually! I think that's what my teacher wants me to be doing, but I enjoy the other method better so far, haha.

Answer 17

yes of course
but we can still do it this way
since cosine and sine are points on the unit circle, we always have
\[\cos^2(x)+\sin^2(x)=1\] so
\[\sin^2(x)=1-\cos^2(x)\] or
\[\sin(x)=\pm\sqrt{1-\cos^2(x)}\]

Answer 18

now lets replace \(\cos(x)\) by \(-\frac{2}{5}\) and you get
\[\sin(x)=\pm\sqrt{1-(\frac{2}{5})^2}\] or
\[\sin(x)=\pm\sqrt{1-\frac{4}{25}}=\pm\sqrt{\frac{21}{25}}\]

Answer 19

since you know \(\sqrt{25}=5\) this is just
\[\sin(x)=\pm\frac{\sqrt{21}}{5}\]

Answer 20

that way is horrid...

Answer 21

you still have to decide whether it is plus or minus
in this case it is minus
yes, this is more annoying
i can show you why there are exactly the same if you like

Answer 22

Please, I'd appreciate all the help I can get!

Answer 23

that is the triangle i drew originally
note that the hypotenuse is 5, but if you are at a point on the unit circle, the hypotenuse is always 1, but the ratios are all the same

Answer 24

so you can say cosine is \(\frac{2}{5}\) as a point on the unit circle, or you can say it is \(\frac{2}{5}\) as in "adjacent over hypotenuse" it is all the same

Answer 25

but for finding the other side without the fraction arithmetic, i find it easier just to use pythagoras and then divide by the hypotenuse
don't forget you still have to determine whether your answer is positive or negative in either method

Answer 26

damn i made a typo in my picture, let me redraw it

Answer 27

radio is 2 to 5
bad picture earlier, sorry if it confused you

Answer 28

and I can figure out whether it is positive or negative by using

Answer 29

no i understood what you meant! that went right over my head as well hahaha

Answer 30

you are told that tangent is positive, and also you know cosine is negative
since tangent is sine over cosine, that means that sine must be negative as well.
no need for memorizing that chart