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stich3 [128]
3 years ago
14

given that sin theta= 1/4, 0 is less than theta but less than pi/2, what is the exact value of cos theta

Mathematics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer:

\cos{\theta} = \frac{\sqrt{15}}{4}

Step-by-step explanation:

For any angle \theta, we have that:

(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1

Quadrant:

0 \leq \theta \leq \frac{\pi}{2} means that \theta is in the first quadrant. This means that both the sine and the cosine have positive values.

Find the cosine:

(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1

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

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

(\cos{\theta})^{2} = 1 - \frac{1}{16}

(\cos{\theta})^{2} = \frac{16-1}{16}

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

\cos{\theta} = \pm \sqrt{\frac{15}{16}}

Since the angle is in the first quadrant, the cosine is positive.

\cos{\theta} = \frac{\sqrt{15}}{4}

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