Question

Find out the values of angle $150°$ for all the six trigonometric ratios.

Answer 1

Step 1: Compute the $\text{sine}$ and $\mathrm{cosine}$ of the given angle.

In the question, the measure of an angle $150°$ is given.

Compute the $\text{sine}$ of the given angle.

We know that, $\sin \left(180°-\theta \right)=\sin \left(\theta \right)$.

So,

$$\begin{gathered} \sin (150°)=\sin \left(\right(180°-30°\left)\right) \\ \Rightarrow \sin (150°)=\sin (30°) \\ \Rightarrow \sin (150°)=\frac{1}{2} \end{gathered}$$

Compute the $\text{cosine}$ of the given angle.

We know that, $\cos \left(180°-\theta \right)=-\cos \left(\theta \right)$.

So,

$$\begin{gathered} \cos (150°)=\cos (150°-30°) \\ \Rightarrow \cos (150°)=-\cos (30°) \\ \Rightarrow \cos (150°)=-\frac{\sqrt{3}}{2} \end{gathered}$$

Step 2: Compute the $\text{tangent}$ and $\mathrm{cotangent}$ of the given angle.

Compute the $\text{tangent}$ of the given angle.

We know that, $\tan \left(180°-\theta \right)=-\tan \left(\theta \right)$.

So,

$$\begin{gathered} \tan (150°)=\tan (150°-30°) \\ \Rightarrow \tan (150°)=-\tan (30°) \\ \Rightarrow \tan (150°)=-\frac{1}{\sqrt{3}} \end{gathered}$$

Compute the $\text{cotangent}$ of the given angle.

We know that, $cot\left(180°-\theta \right)=-cot\left(\theta \right)$.

So,

$$\begin{gathered} cot(150°)=cot(150°-30°) \\ \Rightarrow cot(150°)=-cot(30°) \\ \Rightarrow cot(150°)=-\sqrt{3} \end{gathered}$$

Step 3: Compute the $\text{secant}$ and $\text{cosecant}$ of the given angle.

Compute the $\text{secant}$ of the given angle.

We know that, $sec\left(180°-\theta \right)=-sec\left(\theta \right)$.

So,

$$\begin{gathered} sec(150°)=sec(150°-30°) \\ \Rightarrow sec(150°)=-sec(30°) \\ \Rightarrow sec(150°)=-\frac{2}{\sqrt{3}} \end{gathered}$$

Compute the $\text{cosecant}$ of the given angle.

We know that, $\cos ec\left(180°-\theta \right)=\cos ec\left(\theta \right)$.

So,

$$\begin{gathered} \cos ec(150°)=cosec(150°-30°) \\ \Rightarrow \cos ec(150°)=cosec(30°) \\ \Rightarrow \cos ec(150°)=2 \end{gathered}$$

Hence, the values of trigonometric ratios are

$$\begin{gathered} \sin \left({150}^{\circ }\right)=\frac{1}{2} \\ \cos \left({150}^{\circ }\right)=-\frac{\sqrt{3}}{2} \\ \tan \left({150}^{\circ }\right)=-\frac{1}{\sqrt{3}} \\ c\mathrm{os}ec\left({150}^{\circ }\right)=2 \\ sec\left({150}^{\circ }\right)=-\frac{2}{\sqrt{3}} \\ cot\left({150}^{\circ }\right)=-\sqrt{3} \end{gathered}$$