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

Trigonometry
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Okay, let's make that drawing again, though following @hartnn 's advice, I'll do it properly this time ;) DRAW!!! |dw:1360757734426:dw|
Cosine is adj/hyp, right?
yes

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Other answers:

So, we have cosθ = -(5/9) We place the 5 and the 9, accordingly... |dw:1360757883210:dw| (Stop me when you don't understand)
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 :)
\[\cos ^2(\theta)+\sin^2(\theta)=1\]
\[\sin^2(\theta) = 1-\cos^2(\theta)\]
\[\sqrt{56}\]
\[\sin^2(\theta)=(\frac{- 5 }{ 9 })^2 -1\]
i prefer the method provided by @PeterPan, because it shows you the inside out
So, we can put sqrt(56) on here...|dw:1360758287438:dw| And you can easily get the other trig functions. Be careful though...
@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
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
@savac REMEMBER THIS |dw:1360758381571:dw|
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 :)
So back to this... |dw:1360758603756:dw| Can you now determine the rest of the trig functions of this theta?
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
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
thats what i thought it was
actually since theta is in 2nd quadrant, that is its measure is between 90 to 180, the triangle will be , |dw:1360722028091:dw|
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 :)
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.
Identities are harder to swallow, and they taste kinda awful too :(

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