Question

find the value of the sine, cosine, tangent function for angle theta

Answer 1

We can draw the following picture:

The sine function is given by

$\begin{gathered} \sin \theta=\frac{Opposite}{hypotenuse} \\ \sin \theta=\frac{3}{\sqrt[]{13}} \end{gathered}$

Now, we know that

$\begin{gathered} \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}\times1 \\ \text{and we can write 1 as } \\ 1=\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \text{then} \\ \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \end{gathered}$

but square root of 13 times square root of 13 is equal to 13, I mean

$\sqrt[]{13}\times\sqrt[]{13}=13$

then, we have

$\begin{gathered} \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \frac{3}{\sqrt[]{13}}=\frac{3\sqrt[]{13}}{13} \end{gathered}$

then, an equivalent answer is

$\sin \theta=\frac{3\sqrt[]{13}}{13}$

the cosine function is given by

$\begin{gathered} \cos \theta=\frac{Adjacent}{\text{hypotenuse}} \\ \text{cos}\theta=\frac{2}{\sqrt[]{13}} \end{gathered}$

this answer can be rewritten as

$\begin{gathered} \text{cos}\theta=\frac{2}{\sqrt[]{13}}\times1 \\ \text{cos}\theta=\frac{2}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \text{cos}\theta=\frac{2\sqrt[]{13}}{13} \end{gathered}$

and the tangent funcion is given by

$\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \tan \theta=\frac{3}{2} \end{gathered}$