# If csc theta=4/3, what is the sin, cos, tan, sec, and cot?

Mar 27, 2018

See below.

#### Explanation:

Instead of using formulas, it'd be easier to solve it geometrically, with a right triangle.

Since $\csc \theta = \frac{1}{\sin} \theta = \text{hypotenuse"/"opposite} = \frac{c}{a} = \frac{4}{3}$, this means that $a$ and $c$ are multiples of $3$ and $4$, respectively.

In other words, we have $c = 4 k$ and $a = 3 k$, for a real number $k$.
By the Pythagorean theorem, $b = \sqrt{{c}^{2} - {a}^{2}} = \sqrt{16 {k}^{2} - 9 {k}^{2}} = \sqrt{7} \cdot k$.

Finally, for trigonometric functions :

$\sin \theta = \text{opposite"/"hypotenuse} = \frac{a}{c} = \frac{3}{4}$
$\cos \theta = \text{adjacent"/"hypotenuse} = \frac{b}{c} = \frac{\sqrt{7}}{4}$

$\tan \theta = \text{opposite"/"adjacent} = \frac{a}{b} = \frac{3}{\sqrt{7}}$
$\cot \theta = \frac{1}{\tan} \theta = \frac{b}{a} = \frac{\sqrt{7}}{3}$

$\sec \theta = \text{hypotenuse"/"adjacent} = \frac{c}{b} = \frac{4}{\sqrt{7}}$.

Apr 2, 2018

As below.

#### Explanation:

$\csc \theta = \frac{4}{3}$

$\sin \theta = \frac{1}{\csc} \theta = \frac{1}{\frac{4}{3}} = \frac{3}{4}$

${\cos}^{2} \theta = 1 - {\sin}^{2} \theta = 1 - \frac{9}{16} = \frac{7}{16}$

$\cos \theta = \pm \frac{\sqrt{7}}{4}$

$\sec \theta = \frac{1}{\cos} \theta = \pm \frac{4}{\sqrt{7}}$

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \pm \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \pm \frac{3}{\sqrt{7}}$

$\cot \theta = \frac{1}{\tan} \theta = \pm \frac{\sqrt{7}}{3}$