# Find the exact value of each of the remaining trigonometric functions of theta.

$cos\theta=\frac{24}{25}\ ,\ {270} ^\circ<\theta<{360} ^\circ$

– Part (a) – $sin\theta=?$

– Part (b) – $tan\theta=?$

– Part (c) – $sec\theta=?$

– Part (d) – $csc\theta=?$

– Part (e) – $cot\theta=?$

The aim of the article is to find the value of trigonometric functions of the Right-Angled triangleThe basic concept behind this article is the Right-Angled Triangle and the Pythagorean Identity.

A triangle is called Right-Angled Triangle if it contains one internal angle of ${90}^\circ$ and the other two internal angles sum up with the Right Angle to complete ${180}^\circ$. The horizontal side of the Right Angle is called the Adjacent, and the Vertical Side is called the Opposite.

The Pythagorean Identity for the Right-Angled Triangle is expressed as follows:

$\sin^2\theta+\cos^2\theta=1$

This is true for all values of angles $\theta$.

Given that:

$cos\theta=\frac{24}{25}\ ,\ {270}^\circ<\theta<{360}^\circ$

The given range of angle represents that the angle $\theta$ lies in the $4^{th}$ quadrant.

Part (a) – $sin\theta=?$

As per the Pythagorean Identity, we know that:

$\sin^2\theta+{\ \cos}^2\theta=1$

$sin\theta\ =\ \sqrt{1-\cos^2\theta}$

Substituting the value of $cos\theta=\dfrac{24}{25}$:

$sin\theta=\sqrt{1-\left(\frac{24}{25}\right)^2}$

$sin\theta=\sqrt{\frac{625-576}{625}}$

$sin\theta=\sqrt{\frac{49}{625}}$

$sin\theta=\pm\frac{7}{25}$

Since the angle $\theta$ lies in $4^{th}$ quadrant, the $sine$ function will be negative:

$sin\theta=-\frac{7}{25}$

Part (b) – $tan\theta=?$

We know that for the Right-Angled Triangle:

$tan\theta=\frac{sin\theta}{cos\theta}$

Substituting the value of $sin\theta$ and $cos\theta$ in the above equation:

$tan\theta=\frac{-\dfrac{7}{25}}{\dfrac{24}{25}}$

$tan\theta=-\frac{7}{25}\times\frac{25}{24}$

$tan\theta=-\frac{7}{24}$

Part (c) – $sec\theta=?$

We know that for the Right-Angled Triangle:

$sec\theta=\frac{1}{cos\theta}$

Substituting the value $cos\theta$ in the above equation:

$sec\theta=\frac{1}{\dfrac{24}{25}}$

$sec\theta=\frac{25}{24}$

Part (d) – $csc\theta=?$

We know that for the Right-Angled Triangle:

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

Substituting the value $sin\theta$ in the above equation:

$csc\theta=\frac{1}{-\dfrac{7}{25}}$

$csc\theta=-\frac{25}{7}$

Part (e) – $cot\theta=?$

We know that for the Right-Angled Triangle:

$cot\theta=\frac{1}{tan\theta}$

Substituting the value $tan\ \theta$ in the above equation:

$cot\theta=\frac{1}{-\dfrac{7}{24}}$

$cot\theta=-\frac{24}{7}$

## Numerical Result

Part (a) – $sin\ \theta\ =\ -\ \dfrac{7}{25}$

Part (b) – $tan\ \theta\ =\ -\ \dfrac{7}{24}$

Part (c) – $sec\ \theta\ =\ \dfrac{25}{24}$

Part (d) – $csc\ \theta\ =\ -\ \dfrac{25}{7}$

Part (e) – $cot\ \theta\ =\ -\ \dfrac{24}{7}$

## Example

Calculate the value for the following trigonometric functions if:

$cos\ \theta\ =\ \frac{3}{5}\ ,\ {90}^\circ\ <\ \theta\ <\ {180}^\circ$

Part (a) – $sin\ \theta\ =\ ?$

Part (b) – $tan\ \theta\ =\ ?$

Solution

Given that:

$cos\ \theta\ =\ \frac{3}{5}\ ,\ {90}^\circ\ <\ \theta\ <{\ 180}^\circ$

The given range of angle represents that the angle $\theta$ lies in the $2^{nd}$ quadrant.

Part (a) – $sin\ \theta\ =\ ?$

As per the Pythagorean Identity, we know that:

$\sin^2\ \theta+{\ \cos}^2\ \theta\ =\ 1$

$sin\theta\ =\ \sqrt{1\ -{\cos}^2\ \theta}$

Substituting the value of $cos\ \theta\ =\ \dfrac{3}{5}$:

$sin\ \theta\ =\ \sqrt{1\ -{\ \left(\frac{3}{5}\right)}^2}$

$sin\ \theta\ =\ \sqrt{\frac{25\ -\ 9}{25}}$

$sin\ \theta\ =\ \sqrt{\frac{16}{25}}$

$sin\ \theta\ =\ \pm\ \frac{4}{5}$

Since the angle $\theta$ lies in the $2^{nd}$ quadrant, the $sine$ function will be positive:

$sin\ \theta\ =\ \ \frac{4}{5}$

Part (b) – $tan\ \theta\ =\ ?$

We know that for the Right-Angled Triangle:

$tan\ \theta\ =\ \frac{sin\ \theta}{cos\ \theta}$

Substituting the value of $sin\ \theta$ and $cos\ \theta$ in the above equation:

$tan\ \theta\ =\ \frac{\ \dfrac{4}{5}}{\dfrac{3}{5}}$

$tan\ \theta\ =\ \frac{4}{5}\ \times\ \frac{5}{3}$

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