Question

Given the quadrant of theta in standard position and a trigonometric function value of theta , find the exact value for the indicated function. I, cos theta = 0.25; tan theta

Answer 1

Answer:

$\sqrt{15}$

Step-by-step explanation:

$\cos(\theta)=\frac{1}{4}$ is given.

My attempt will to be to use the Pythagorean Identity:

$\cos^2(\theta)+\sin^2(\theta)=1$

$(\frac{1}{4})^2+\sin^2(\theta)=1$

$\frac{1}{16}+\sin^2(\theta)=1$

Subtract 1/16 on both sides:

$\sin^2(\theta)=\frac{15}{16}$

We have the following quotient identity:

$\frac{\sin^2(\theta)}{\cos^2(\theta)}=\tan^2(\theta)$

$\frac{\sin^2(\theta)}{\cos^2(\theta)}=\frac{\frac{15}{16}}{\frac{1}{16}}$

Multiply the numerator and denominator by 16. This is multiplying the fraction by 1 so this doesn't change the value of the fraction.

$\tan^2(\theta)=\frac{15}{1}$

$\tan^2(\theta)=15$

Square root both sides:

$\tan(\theta)=\pm \sqrt{15}$

Since we are in the first quadrant then all 6 trigonometric functions are positive there so the answer is $\sqrt{15}$.