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Cosine is adj/hyp, right?

yes

i understand

Now, what's the length of this missing side... |dw:1360758005338:dw|
(Hint: Pythagorean theorem)

\[\sqrt{106}\] right?

Remember, 9 is your hypotenuse
When you have
a^2 + b^2 = c^2
9 is your c :)

\[\cos ^2(\theta)+\sin^2(\theta)=1\]

\[\sin^2(\theta) = 1-\cos^2(\theta)\]

\[\sqrt{56}\]

\[\sin^2(\theta)=(\frac{- 5 }{ 9 })^2 -1\]

yes that is what i said

i said i prefer the other method to know the inside out

we are allowed to draw on the test

although the identity was proven using pyth.

whats that?

ACTS!!!!
|dw:1360758433801:dw|

ohhhh thats smart

savac PeterPan is doing a great job

@mathsmind
Thanks for the kudos!
Much appreciated :)

give me a mint

min

:D
I was about to say something clever, but you beat me to it :)

REMEMBER
In this particular quadrant, only sine (and cosecant) are positive

lol sorry running low on sleep

ok i got sin=\[\sqrt{56}\]/9
tan=-\[\sqrt{56}\]/5
csc=9/\[\sqrt{56}\]
sec=-9/5
cot=-5/\[\sqrt{56}\]

What about cos?

cos=-5/9

And there you have it :)

that was rright?

Well yeah, unless you want to make it
\[\huge \sqrt{56}=2\sqrt{14}\]

wow i surprised myself. lol thank you very much

thats what i thought it was

Looks more complicated.

but the way its solved and answers are correct.

sin and cosec will be positive, all other ratios will be negative.

Yay :)

Identities are harder to swallow, and they taste kinda awful too :(