Question

The point (-4,10) is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

Answer 1

Hint: In this question, we want to determine the exact values of the six trigonometric functions of the angle in a standard position to the point (-4,10). For that, we will apply the formula $r = \sqrt {{x^2} + {y^2}} $ and calculate r. Here, x and y are the coordinates of the given point. The r will be the hypotenuse, the opposite side will be y, and the adjacent side will be x. Then find all the trigonometric ratios.
The six trigonometric functions of the angles are as below:
$\sin \theta = \dfrac{{opposite}}{{hypotenuse}}$
$\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}$
$\tan \theta = \dfrac{{opposite}}{{adjacent}}$
$\cos ec\theta = \dfrac{{hypotenuse}}{{opposite}}$
$\sec \theta = \dfrac{{hypotenuse}}{{adjacent}}$
$\cot \theta = \dfrac{{adjacent}}{{opposite}}$

Complete step-by-step solution:
In this question, the given point P(-4,10) is on the terminal side of an angle in standard position.
Here, let x will be -4, and y will be 10.
Now, apply the formula:
 $ \Rightarrow r = \sqrt {{x^2} + {y^2}} $
Substitute the value of x and y in the above equation.
$ \Rightarrow r = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( {10} \right)}^2}} $
That is equal to,
$ \Rightarrow r = \sqrt {16 + 100} $
Let us add the right-hand side.
$ \Rightarrow r = \sqrt {116} $
Let us apply the square root on the right-hand side.
$ \Rightarrow r = 2\sqrt {29} $
The $r = 2\sqrt {29} $ will be the hypotenuse, the opposite side will be y=10, and the adjacent side will be x=-4.
Now, let us find the value of the sine function.
$ \Rightarrow \sin \theta = \dfrac{{opposite}}{{hypotenuse}}$
Therefore,
$ \Rightarrow \sin \theta = \dfrac{{10}}{{2\sqrt {29} }}$
$ \Rightarrow \sin \theta = \dfrac{5}{{\sqrt {29} }}$
Multiply the numerator and the denominator by $\sqrt {29} $ to convert them into the standard form.
 $ \Rightarrow \sin \theta = \dfrac{{5\sqrt {29} }}{{29}}$
Now, let us find the value of the cosine function.
$ \Rightarrow \cos \theta = \dfrac{{adjacent}}{{hypotenuse}}$
Therefore,
$ \Rightarrow \cos \theta = \dfrac{{ - 4}}{{2\sqrt {29} }}$
$ \Rightarrow \cos \theta = - \dfrac{2}{{\sqrt {29} }}$
Multiply the numerator and the denominator by $\sqrt {29} $ to convert them into the standard form.
 $ \Rightarrow \cos \theta = - \dfrac{{2\sqrt {29} }}{{29}}$
Now, let us find the value of the tangent function.
$ \Rightarrow \tan \theta = \dfrac{{opposite}}{{adjacent}}$
Therefore,
$ \Rightarrow \tan \theta = \dfrac{{10}}{{ - 4}}$
$ \Rightarrow \tan \theta = - 2.5$
Now, let us find the value of the cosecant function.
$ \Rightarrow \cos ec\theta = \dfrac{{hypotenuse}}{{opposite}}$
Therefore,
$ \Rightarrow \cos ec\theta = \dfrac{{2\sqrt {29} }}{{10}}$
$ \Rightarrow \cos ec\theta = \dfrac{{\sqrt {29} }}{5}$
Now, let us find the value of the secant function.
$ \Rightarrow \sec \theta = \dfrac{{hypotenuse}}{{adjacent}}$
Therefore,
$ \Rightarrow \sec \theta = \dfrac{{2\sqrt {29} }}{{ - 4}}$
$ \Rightarrow \sec \theta = - \dfrac{{\sqrt {29} }}{2}$
Now, let us find the value of the cotangent function.
$ \Rightarrow \cot \theta = \dfrac{{adjacent}}{{opposite}}$
Therefore,
$ \Rightarrow \cot \theta = \dfrac{{ - 4}}{{10}}$
$ \Rightarrow \cot \theta = - \dfrac{2}{5}$

Note: The point we end up at is called the terminal point P(x,y). The vertex is always placed at the origin and one ray is always placed on the positive x-axis. This ray is called the initial side of the angle. The other ray is called the terminal side of the angle. This position of an angle is called the standard position. An angle is formed by two rays that have a common end-point. The common end-point is called the vertex. The coordinate plane is divided into four regions or quadrants. An angle can be located in any of the four quadrants, depending on which quadrant contains its terminal side.