Question

An equation of the terminal side of an angle θ in standard position is given with a restriction on x. Sketch the least positive such angle θ , and find the values of the six trigonometric functions of θ . 12x + 5y = 0 , x ≥ 0 .

Accepted Answer

Rewrite the given equation of the terminal side in slope-intercept form to understand the line better. Starting with $$12x + 5y = 0$$, solve for $$y to $$get $$y = -\frac{12}{5}x. $$Recognize that the terminal side of angle $$	heta$$ lies along the line $$y = -\frac{12}{5}x$$ with the restriction $$x \geq 0. $$This means the terminal side is in the fourth quadrant or on the positive x-axis. Find the reference angle $$\alpha by $$considering the slope as the tangent of the angle the line makes with the positive x-axis. Use $$	an(\alpha) = \left| -\frac{12}{5} ight| = \frac{12}{5}$$, then find $$\alpha = \arctan\left(\frac{12}{5}ight). $$Determine the least positive angle $$	heta in $$standard position. Since the line is in the fourth quadrant (because $$x \geq 0$$ and $$y is $$negative), calculate $$	heta = 2\pi - \alpha (in $$radians) or $$360^\circ - \alpha (in $$degrees). Calculate the six trigonometric functions of $$	heta by $$first choosing a point on the terminal side that satisfies the line equation and $$x \geq 0. $$For example, use $$x = 5$$, then $$y = -12. $$Compute the hypotenuse $$r = \sqrt{x^2$ + y^2$} = \sqrt{5^2$ + (-12)^2$}. $$Then use the definitions: $$\sin 	heta = \frac{y}{r}$$, $$\cos 	heta = \frac{x}{r}$$, $$	an 	heta = \frac{y}{x}$$, $$\csc 	heta = \frac{r}{y}$$, $$\sec 	heta = \frac{r}{x}$$, and $$\cot 	heta = \frac{x}{y}.$$

An equation of the terminal side of an angle θ in standard position is given with a restriction on x. Sketch the least positive such angle θ , and find the values of the six trigonometric functions of θ . 12x + 5y = 0 , x ≥ 0 .

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Key Concepts

Standard Position of an Angle

Equation of a Line and Its Relation to an Angle

Six Trigonometric Functions