Maribel Ali

2023-03-05

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

kizintefevd3

$\mathrm{sin}\theta =\frac{3}{8}$
By rearranging the Pythagorean theorem, we can calculate the adjacent side, b, since we already have the side opposite to $\theta$, $3$, and the hypotenuse, $8$.
${a}^{2}+{b}^{2}={c}^{2}$
${b}^{2}={c}^{2}-{a}^{2}$
${b}^{2}={8}^{2}-{3}^{2}$
${b}^{2}=64-9$
${b}^{2}=55$
$b=-\sqrt{55}$
From this, we can determine that the side that is next to it has a length of $-\sqrt{55}$ units, since the x-axis is negative in the second quadrant. Therefore, we can infer that $\mathrm{cos}\theta =-\frac{\sqrt{55}}{8}$ and $\mathrm{tan}\theta =-\frac{3}{\sqrt{55}}$
Now, using the reciprocal identities, we must take the reciprocals of forces, sine, and cosine.
$\mathrm{csc}\theta =\frac{1}{\mathrm{sin}\theta }$
$\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }$
$\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }$
So,
$\mathrm{csc}\theta =\frac{8}{3},\mathrm{sec}\theta =-\frac{8}{\sqrt{55}}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\mathrm{cot}\theta =-\frac{\sqrt{55}}{3}$
In conclusion, the six trigonometric ratios are:
$\mathrm{sin}\theta =\frac{3}{8}$
$\mathrm{cos}\theta =-\frac{\sqrt{55}}{8}$
$\mathrm{tan}\theta =-\frac{3}{\sqrt{55}}$
$\mathrm{csc}\theta =\frac{8}{3}$
$\mathrm{sec}\theta =-\frac{8}{\sqrt{55}}$
$\mathrm{cot}\theta =-\frac{\sqrt{55}}{3}$

Nhluvukoj6m

$\mathrm{sin}\theta =\frac{3}{8}\to \left(1\right)$
$•\mathrm{csc}\theta =\frac{1}{\mathrm{sin}\theta }=\frac{8}{3}\to \left(2\right)$
$•\mathrm{cos}\theta =±\sqrt{1-{\mathrm{sin}}^{2}\theta }$
$××x=-\sqrt{1-\frac{9}{64}}\phantom{\rule{1ex}{0ex}}\text{}$
$××x=-\sqrt{\frac{55}{64}}$
$⇒\mathrm{cos}\theta =-\frac{\sqrt{55}}{8}\to \left(3\right)$
$•\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }=-\frac{8}{\sqrt{55}}\to \left(4\right)$
$•\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }=-\frac{8}{\sqrt{55}}\to \left(4\right)$
$•\mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }$
$××=\frac{3}{8}×-\frac{8}{\sqrt{55}}=-\frac{3}{\sqrt{55}}\to \left(5\right)$
$•\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }=-\frac{\sqrt{55}}{3}\to \left(6\right)$

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