If sin of theta equals 3/8 and theta is in quadrant II. what are cos, tan, csc, cot, and sec of theta?

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7
Noah G Share
Apr 14, 2016

See explanation below.

Explanation:

$\sin \theta = \frac{3}{8}$, $\theta$ in quadrant $I I$

Imagine a right triangle being drawn on the cartesian plane, as in the following example.

Since sin = opposite/hypotenuse, the side opposite to $\theta$ is 3 and the hypotenuse is $8$, we can rearrange our pythagorean theorem to find the adjacent side, b.

${a}^{2} + {b}^{2} = {c}^{2}$

${b}^{2} = {c}^{2} - {a}^{2}$

${b}^{2} = {8}^{2} - {3}^{2}$

${b}^{2} = 64 - 9$

${b}^{2} = 55$

$b = - \sqrt{55}$

So, now we know that the adjacent side measures $- \sqrt{55}$ units (since the x axis is negative in quadrant II) in length. Thus we can deduce that $\cos \theta = - \frac{\sqrt{55}}{8}$ and $\tan \theta = - \frac{3}{\sqrt{55}}$

Now, for $\csc \theta , \sec \theta \mathmr{and} \cot \theta$, we must apply the reciprocal identities.

$\csc \theta = \frac{1}{\sin} \theta$

$\sec \theta = \frac{1}{\cos} \theta$

$\cot \theta = \frac{1}{\tan} \theta$

Therefore, $\csc \theta = \frac{8}{3} , \sec \theta = - \frac{8}{\sqrt{55}} \mathmr{and} \cot \theta = - \frac{\sqrt{55}}{3}$

To summarize, the six trigonometric ratios, we get:

$\sin \theta = \frac{3}{8}$

$\cos \theta = - \frac{\sqrt{55}}{8}$

$\tan \theta = - \frac{3}{\sqrt{55}}$

$\csc \theta = \frac{8}{3}$

$\sec \theta = - \frac{8}{\sqrt{55}}$

$\cot \theta = - \frac{\sqrt{55}}{3}$

Hopefully this helps!

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Jim G. Share
May 24, 2017

$\text{see explanation}$

Explanation:

$\sin \theta = \frac{3}{8} \to \left(\textcolor{red}{1}\right)$

• csctheta=1/sintheta=8/3to(color(red)(2))

• costheta=+-sqrt(1-sin^2theta)

$\textcolor{w h i t e}{\times \times x} = - \sqrt{1 - \frac{9}{64}} \leftarrow \text{ negative value}$

$\textcolor{w h i t e}{\times \times x} = - \sqrt{\frac{55}{64}}$

$\Rightarrow \cos \theta = - \frac{\sqrt{55}}{8} \to \left(\textcolor{red}{3}\right)$

• sectheta=1/costheta=-8/sqrt55to(color(red)(4))

• tantheta=(sintheta)/(costheta)

$\textcolor{w h i t e}{\times \times} = \frac{3}{8} \times - \frac{8}{\sqrt{55}} = - \frac{3}{\sqrt{55}} \to \left(\textcolor{red}{5}\right)$

• cottheta=1/tantheta=-sqrt55/3to(color(red)(6))

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