\[cos\theta=\frac{24}{25}\ ,\ {270} ^\circ<\theta<{360} ^\circ\]
– 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 triangle. The 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 $\theta$.
Expert Answer
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:
\[tan\ \theta\ =\ \frac{\ \dfrac{4}{5}}{\dfrac{3}{5}} \]
\[tan\ \theta\ =\ \frac{4}{5}\ \times\ \frac{5}{3} \]
\[tan\ \theta\ =\ \frac{4}{3} \]