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RideAnS [48]
4 years ago
7

Find the exact values of the six trigonometric functions of θ if θ is in standard position and the terminal side of θ is in the

specified quadrant and satisfies the given condition.
III; parallel to the line
5y − 2x + 1 = 0
Mathematics
2 answers:
mixer [17]4 years ago
7 0
5y - 2x + 1 = 0.
5y = 2x-1.

y = 2/5 x - 2/5

this tells us that tan(theta) = 2/5
sin(theta)/sqrt(1-sin^2(theta)) = 2/5
x/sqrt(1-x^2)=2/5
x^2/(1-x^2)=4/25
25x^2=4-4x^2
29x^2=4
x^2=4/29
x = -2/sqrt(29) because quadrant III
cos(x) = sqrt(1-4/29) = -5/sqrt(29)


tan(theta)=2/5
cos(theta)=-5/sqrt(29)
sin(theta)=-2/sqrt(29)
cot(theta)=5/2
sec(theta)=-sqrt(29)/5
csc(theta)=-sqrt(29)/2
Pani-rosa [81]4 years ago
7 0
So the equation of the line is 5y -2x + 1 = 0.

Let's put that in standard form (by solving for y):

5y -2x + 1 = 0
5y = 2x - 1 ~ we added 2x and subtracted 1 from both sides
y =  \frac{2}{5}x - \frac{1}{5} ~ divide both sides by 5

So now we have it in the standard form which is y = mx + b.

The m gives us the slope. Recall that slope is rise over run.

So the slope is \frac{2}{5}, which means that you could draw a triangle with a rise of 2 and a run of 5.

That's almost enough information to find the trig values, but we still need to know the hypotenuse. Well, thanks to the magic of the Pythagorean Theorem:

hyp =  \sqrt{2^2 + 5^2} =  \sqrt{4+25} =  \sqrt{29}

Alright, now that we know all three sides, we can find the values.

The "rise" (vertical) or opposite: 2
The "run" (horozontial) or adjacent: 5
The hypotenuse: \sqrt{29}

Remember SOHCAHTOA. In order from that mnemonic: 

sin(θ) = opposite over hypotenuse = \frac{2}{\sqrt{29}}

cos(θ) = adjacent over hypotenuse = \frac{5}{\sqrt{29}}

tan(θ) = opposite over adjacent = \frac{2}{5}

Now the others (cosecant, secant, and cotangent) are just the reciprocals (upside-downs) of the others. Remember that cosecant comes first (since sine comes first in the "usuals".

csc(θ) = \frac{\sqrt{29}}{2}

sec(θ) = \frac{\sqrt{29}}{5}

cot(θ) = \frac{5}{2}

We don't have to 'solve' (approximate) that square root in any of the answers since the problem asked for the exact solution.
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