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=(Opposite)/(hypotenuse) \\ \sin \theta=\frac{3}{\sqrt[]{13}} \end{gathered}`

Now, we know that

`\begin{gathered} \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}*1 \\ \text{and we can write 1 as } \\ 1=\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \text{then} \\ \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}*\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}*\sqrt[]{13}=13`

then, we have

`\begin{gathered} \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}*\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}}*1 \\ \text{cos}\theta=\frac{2}{\sqrt[]{13}}*\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=(opposite)/(adjacent) \\ \tan \theta=(3)/(2) \end{gathered}`