Use the given value and the trigonometric identities to find the remaining trigonometric functions of the angle.
cos θ = − 5/9, sin θ > 0

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- anonymous

- schrodinger

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- anonymous

Okay, let's make that drawing again, though following @hartnn 's advice, I'll do it properly this time ;)
DRAW!!!
|dw:1360757734426:dw|

- anonymous

Cosine is adj/hyp, right?

- anonymous

yes

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## More answers

- anonymous

So, we have cosθ = -(5/9)
We place the 5 and the 9, accordingly...
|dw:1360757883210:dw|
(Stop me when you don't understand)

- anonymous

i understand

- anonymous

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

- anonymous

\[\sqrt{106}\] right?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

\[\sqrt{56}\]

- anonymous

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

- anonymous

i prefer the method provided by @PeterPan, because it shows you the inside out

- anonymous

So, we can put sqrt(56) on here...|dw:1360758287438:dw|
And you can easily get the other trig functions. Be careful though...

- anonymous

@mathsmind
You do need to do this to illustrate the pythagorean identities you posted. Though if you're not allowed to draw, better know the pythagorean (and other) identities by heart!!!
:D

- anonymous

yes that is what i said

- anonymous

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

- anonymous

we are allowed to draw on the test

- anonymous

@savac
REMEMBER THIS
|dw:1360758381571:dw|

- anonymous

although the identity was proven using pyth.

- anonymous

whats that?

- anonymous

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

- anonymous

ohhhh thats smart

- anonymous

savac PeterPan is doing a great job

- anonymous

@mathsmind
Thanks for the kudos!
Much appreciated :)

- anonymous

So back to this...
|dw:1360758603756:dw|
Can you now determine the rest of the trig functions of this theta?

- anonymous

give me a mint

- anonymous

min

- anonymous

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

- anonymous

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

- anonymous

lol sorry running low on sleep

- anonymous

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

- anonymous

What about cos?

- anonymous

cos=-5/9

- anonymous

And there you have it :)

- anonymous

that was rright?

- anonymous

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

- anonymous

wow i surprised myself. lol thank you very much

- anonymous

Just place that right triangle in the correct quadrant, and you can't go wrong :)
You'll notice that the horizontal bit was to the left of the x-axis, so strictly speaking, that's a -5, not a 5 :D

- anonymous

thats what i thought it was

- hartnn

actually since theta is in 2nd quadrant, that is its measure is between 90 to 180, the triangle will be , |dw:1360722028091:dw|

- anonymous

Looks more complicated.

- hartnn

but the way its solved and answers are correct.

- hartnn

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

- anonymous

Yay :)

- hartnn

if you want to solve using diagram, then what i have drawn must be used .
here, its solved using identities, which is altogether different method and doesn't need drawing triangle.

- anonymous

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

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