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adell [148]
4 years ago
5

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

Mathematics
1 answer:
natali 33 [55]4 years ago
3 0
<h3>Answer: Choice B \frac{\sqrt{15}}{4}</h3>

===========================================================

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}\\\\

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move the 9 to the other side.
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I hope this helps you




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krek1111 [17]

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The given quadratic equation is:

-x^{2} +3x+c=0

<h3>What is a quadratic equation?</h3>

Any equation is called a quadratic equation if it has a form ax^{2} +bx+c=0 and a≠0.

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D < 0

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The quadratic equation –x2+3x+c =0 does not have any real solution for c<-9/4.

To get more about quadratic equations visit:

brainly.com/question/1214333

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