Mathematics
emilyfew919
3

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

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(2) Answers
Anaharo0122

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

lidiasefora

So the equation of the line is [latex]5y -2x + 1 = 0[/latex]. Let's put that in standard form (by solving for y): [latex]5y -2x + 1 = 0[/latex] [latex]5y = 2x - 1[/latex] ~ we added 2x and subtracted 1 from both sides [latex]y = \frac{2}{5}x - \frac{1}{5}[/latex] ~ divide both sides by 5 So now we have it in the standard form which is [latex] y = mx + b [/latex]. The [latex] m [/latex] gives us the slope. Recall that slope is rise over run. So the slope is [latex] \frac{2}{5} [/latex], 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: [latex]hyp = \sqrt{2^2 + 5^2} = \sqrt{4+25} = \sqrt{29} [/latex] 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: [latex] \sqrt{29} [/latex] Remember SOHCAHTOA. In order from that mnemonic:  sin(θ) = opposite over hypotenuse = [latex] \frac{2}{\sqrt{29}}[/latex] cos(θ) = adjacent over hypotenuse = [latex] \frac{5}{\sqrt{29}} [/latex] tan(θ) = opposite over adjacent = [latex] \frac{2}{5} [/latex] 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) = [latex] \frac{\sqrt{29}}{2} [/latex] sec(θ) = [latex] \frac{\sqrt{29}}{5} [/latex] cot(θ) = [latex] \frac{5}{2} [/latex] 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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