$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{0}}}\implies -\cfrac{1}{3}$

5 0

2 years ago

Given that sin theta = 1/4, 0→theta→π/2, what<br>. what is the exact value of cos 8?

natali 33 [55]

<h3>Answer: Choice B $\frac{\sqrt{15}}{4}$ </h3>

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Work Shown:

Angle theta is between 0 and pi/2, so this angle is in quadrant Q1.

Square both sides of the given equation

$\sin \theta = \frac{1}{4}\\\\\sin^2 \theta = \left(\frac{1}{4}\right)^2\\\\\sin^2 \theta = \frac{1}{16}$

Then use the pythagorean trig identity to get

$\sin^2 \theta + \cos^2 \theta = 1\\\\\cos^2 \theta = 1-\sin^2 \theta\\\\\cos \theta = \sqrt{1-\sin^2 \theta} \ \ \ \text{cosine is positive in Q1}\\\\\cos \theta = \sqrt{1-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16}{16}-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16-1}{16}}\\\\\cos \theta = \sqrt{\frac{15}{16}}\\\\\cos \theta = \frac{\sqrt{15}}{\sqrt{16}}\\\\\cos \theta = \frac{\sqrt{15}}{4}\\\\$

3 0

2 years ago