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QuestionFind the six trig functions of angle A for the point (−12,−5)(-12,-5). Provide exact values for sin⁡A\sin A, cos⁡A\cos A, tan⁡A\tan A, sec⁡A\sec A, cot⁡A\cot A, and csc⁡A\csc A.

Studdy Solution

STEP 1

Assumptions1. The point on the terminal side of angle A is (-12,-5) in standard position. . We are asked to find the six trigonometric functions of angle A.
3. The six trigonometric functions are sine (sin), cosine (cos), tangent (tan), secant (sec), cotangent (cot), and cosecant (csc).
4. The values of these functions are based on the ratios of the sides of a right triangle formed by the angle in standard position.

STEP 2

In standard position, the x-coordinate of the point represents the length of the adjacent side (or horizontal side) of the right triangle, and the y-coordinate represents the length of the opposite side (or vertical side). The hypotenuse is the distance from the origin (0,0) to the point.Let's denote the x-coordinate as xx, the y-coordinate as yy, and the hypotenuse as rr.

STEP 3

Substitute the given point into the variables xx and yy.
x=−12x = -12y=−5y = -5

STEP 4

We can find the length of the hypotenuse rr using the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
r=x2+y2r = \sqrt{x^2 + y^2}

STEP 5

Substitute the values of xx and yy into the equation to find rr.
r=(−12)2+(−5)2r = \sqrt{(-12)^2 + (-5)^2}

STEP 6

Calculate the value of rr.
r=144+25=169=13r = \sqrt{144 +25} = \sqrt{169} =13

STEP 7

The six trigonometric functions are defined as followssin⁡A=yr\sin A = \frac{y}{r}cos⁡A=xr\cos A = \frac{x}{r}tan⁡A=yx\tan A = \frac{y}{x}sec⁡A=rx\sec A = \frac{r}{x}cot⁡A=xy\cot A = \frac{x}{y}csc⁡A=ry\csc A = \frac{r}{y}

STEP 8

Substitute the values of xx, yy, and rr into the equations to find the values of the trigonometric functions.
sin⁡A=−513\sin A = \frac{-5}{13}cos⁡A=−1213\cos A = \frac{-12}{13}tan⁡A=−5−12=512\tan A = \frac{-5}{-12} = \frac{5}{12}sec⁡A=13−12=−1312\sec A = \frac{13}{-12} = -\frac{13}{12}cot⁡A=−12−5=125\cot A = \frac{-12}{-5} = \frac{12}{5}csc⁡A=13−5=−135\csc A = \frac{13}{-5} = -\frac{13}{5}So, the six trigonometric functions of angle A aresin⁡A=−513\sin A = -\frac{5}{13}cos⁡A=−1213\cos A = -\frac{12}{13}tan⁡A=512\tan A = \frac{5}{12}sec⁡A=−1312\sec A = -\frac{13}{12}cot⁡A=125\cot A = \frac{12}{5}csc⁡A=−135\csc A = -\frac{13}{5}

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