Question

How do you find all six trigonometric functions of $\theta$ if the point $\left( 3,4 \right)$ is on the terminal side of $\theta ?$

Answer 1

The six trigonometric functions are sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec), and cotangent (cot). The point $\left( 3,4 \right)$ lies on the first quadrant. So, the terminal side of the angle $\theta$ lies in the first quadrant. We find the tangent of the angle using the identity $\tan \theta =\dfrac{y}{x}.$ Using this we will find the rest of the trigonometric functions.

Complete step-by-step solution:
Consider the point $\left( 3,4 \right).$
This point lies in the first quadrant where all the trigonometric functions are positive.
Because both the $x$-coordinate and the $y$-coordinate are positive.
Thus, the terminal side of the angle $\theta$ is located in the first quadrant.
We know that the tangent of an angle is the quotient obtained from dividing the $y$-coordinate with the $x$-coordinate.
That is, if the point $\left( 3,4 \right)$ is on the terminal side of the angle $\theta ,$ then the tangent is given by
$\Rightarrow \tan \theta =\dfrac{y}{x}$
And this leads us to the value of the tangent.
That is,
$\Rightarrow \tan \theta =\dfrac{4}{3}$
Now we have the identity that connects tangent and cosine of an angle given as,
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{1+{{\tan }^{2}}\theta }$
Now that we have this identity, we substitute the value of the tangent in this identity to get the value of cosine of the angle $\theta .$
That gives us,
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{1+{{\left( \dfrac{4}{3} \right)}^{2}}}$
We calculate to get,
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{1+\dfrac{{{4}^{2}}}{{{3}^{2}}}}$
And we get,
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{1+\dfrac{16}{9}}$
We take lcm to get,
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{\dfrac{9+16}{9}}=\dfrac{9}{9+16}$ we know, $\dfrac{1}{\left( \dfrac{x}{y} \right)}=\dfrac{y}{x}$
So, we get
$\Rightarrow {{\cos }^{2}}\theta =\dfrac{9}{25}$
This directs us to,
$\Rightarrow \cos \theta =\dfrac{3}{5}$ by taking the square root of the whole equation.
To find the sine of the given angle we use the value of cosine with the help of a familiar identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.$
We get,
$\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta =1-\dfrac{9}{25}$
By simple calculation we get,
$\Rightarrow {{\sin }^{2}}\theta =\dfrac{25-9}{25}=\dfrac{16}{25}$
Taking the square root of the equation,
$\Rightarrow \sin \theta =\dfrac{4}{5}$
Now we can think of finding the cotangent of the angle $\theta$ by taking the reciprocal of the value of the tangent of the angle $\theta .$
Applying the above said identity will give us,
$\Rightarrow \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{3}{4}$
Take the reciprocal of the value of sine to get the value of cosecant,
$\Rightarrow \ cosec\theta =\dfrac{1}{\sin \theta }=\dfrac{5}{4}.$
Find the reciprocal value of cosine to get the value of secant,
$\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }=\dfrac{5}{3}.$

Note: Let us recall some terminologies we have learnt:
Initial side of an angle: The side of an angle from which the rotation begins.
Terminal side of an angle: The side of an angle after rotation.
Also remember that the correct usage of the identities will make the calculation easier.