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Math  /  Trigonometry

QuestionThe point is on the terminal side of an angle in standard position. Find the exact values of the six trigonometric functions of the angle. (312,−215)sin⁡θ=□cos⁡θ=□tan⁡θ=□csc⁡θ=□sec⁡θ=□cot⁡θ=□\begin{array}{l} \left(3 \frac{1}{2},-2 \sqrt{15}\right) \\ \sin \theta=\square \\ \cos \theta=\square \\ \tan \theta=\square \\ \csc \theta=\square \\ \sec \theta=\square \\ \cot \theta=\square \end{array}

Studdy Solution

STEP 1

1. The given point (312,−215)\left(3 \frac{1}{2}, -2 \sqrt{15}\right) is on the terminal side of an angle θ\theta in standard position.
2. The coordinates (x,y) (x, y) are (72,−215) \left( \frac{7}{2}, -2 \sqrt{15} \right) .
3. The hypotenuse r r can be found using the Pythagorean theorem: r=x2+y2 r = \sqrt{x^2 + y^2} .
4. The six trigonometric functions can be defined in terms of x x , y y , and r r .

STEP 2

1. Find the hypotenuse r r .
2. Calculate the sine function sin⁡θ \sin \theta .
3. Calculate the cosine function cos⁡θ \cos \theta .
4. Calculate the tangent function tan⁡θ \tan \theta .
5. Calculate the cosecant function csc⁡θ \csc \theta .
6. Calculate the secant function sec⁡θ \sec \theta .
7. Calculate the cotangent function cot⁡θ \cot \theta .

STEP 3

Find the hypotenuse r r using the Pythagorean theorem:
r=x2+y2 r = \sqrt{x^2 + y^2}
Given x=72 x = \frac{7}{2} and y=−215 y = -2\sqrt{15} ,
r=(72)2+(−215)2 r = \sqrt{\left(\frac{7}{2}\right)^2 + \left(-2\sqrt{15}\right)^2}

STEP 4

Simplify and calculate the hypotenuse r r :
r=(72)2+(−215)2=494+4⋅15=494+60=49+2404=2894=2892=172 r = \sqrt{\left(\frac{7}{2}\right)^2 + \left(-2\sqrt{15}\right)^2} = \sqrt{\frac{49}{4} + 4 \cdot 15} = \sqrt{\frac{49}{4} + 60} = \sqrt{\frac{49 + 240}{4}} = \sqrt{\frac{289}{4}} = \frac{\sqrt{289}}{2} = \frac{17}{2}

STEP 5

Calculate the sine function sin⁡θ \sin \theta :
sin⁡θ=yr=−215172=−41517 \sin \theta = \frac{y}{r} = \frac{-2 \sqrt{15}}{\frac{17}{2}} = \frac{-4 \sqrt{15}}{17}

STEP 6

Calculate the cosine function cos⁡θ \cos \theta :
cos⁡θ=xr=72172=717 \cos \theta = \frac{x}{r} = \frac{\frac{7}{2}}{\frac{17}{2}} = \frac{7}{17}

STEP 7

Calculate the tangent function tan⁡θ \tan \theta :
tan⁡θ=yx=−21572=−4157 \tan \theta = \frac{y}{x} = \frac{-2 \sqrt{15}}{\frac{7}{2}} = \frac{-4 \sqrt{15}}{7}

STEP 8

Calculate the cosecant function csc⁡θ \csc \theta :
csc⁡θ=1sin⁡θ=1−41517=17−415=−17415⋅1515=−171560 \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{-4 \sqrt{15}}{17}} = \frac{17}{-4 \sqrt{15}} = -\frac{17}{4 \sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = -\frac{17 \sqrt{15}}{60}

STEP 9

Calculate the secant function sec⁡θ \sec \theta :
sec⁡θ=1cos⁡θ=1717=177 \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{7}{17}} = \frac{17}{7}

STEP 10

Calculate the cotangent function cot⁡θ \cot \theta :
cot⁡θ=1tan⁡θ=1−4157=7−415=−7415⋅1515=−71560 \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{-4 \sqrt{15}}{7}} = \frac{7}{-4 \sqrt{15}} = -\frac{7}{4 \sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = -\frac{7 \sqrt{15}}{60}
Solution: sin⁡θ=−41517 \sin \theta = -\frac{4 \sqrt{15}}{17} cos⁡θ=717 \cos \theta = \frac{7}{17} tan⁡θ=−4157 \tan \theta = -\frac{4 \sqrt{15}}{7} csc⁡θ=−171560 \csc \theta = -\frac{17 \sqrt{15}}{60} sec⁡θ=177 \sec \theta = \frac{17}{7} cot⁡θ=−71560 \cot \theta = -\frac{7 \sqrt{15}}{60}

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