Question 3, Exercise 8.1

Solutions of Question 3 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the exact value of $\cos 120^{\circ}$ by using $\cos \left(180^{\circ}-60^{\circ}\right)$ and $\cos \left(90^{\circ}+30^{\circ}\right)$.

Solution.

$$\begin{align*} \cos 120^{\circ} & = \cos \left(180^{\circ}-60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*}$$ Also $$\begin{align*} \cos 120^{\circ} & = \cos \left(90^{\circ}+30^{\circ}\right) \\ &= - \sin 30^{\circ}\\ &= -\dfrac{1}{2}. \end{align*}$$

Find the exact value of $\sin 120^{\circ}$ and then $\tan 120^{\circ}$.

Solution.

$$\begin{align*} \sin 120^{\circ} & = \sin \left(180^{\circ}-60^{\circ}\right) \\ &= \sin 60 ^{\circ}\\ &= \dfrac{\sqrt{3}}{2}. \end{align*}$$

Also, we have $$\begin{align*} \cos 120^{\circ} & = \cos \left(180^{\circ}-60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*}$$

Now $$\begin{align*} \tan 120^{\circ} & = \dfrac{\sin 120^{\circ}}{\cos 120^{\circ}} \\ &= \dfrac{\sqrt{3}/2}{-1/2}\\ &= -\sqrt{3}. \end{align*}$$

Find the exact value of $\cos 75^{\circ}$ by using $\cos \left(120^{\circ}-45^{\circ}\right)$.

Solution.

$$\begin{align*} \cos 75^{\circ} & = \cos \left(120^{\circ}-45^{\circ}\right) \\ &= \cos 120^{\circ} \cos 45^{\circ} + \sin 120^{\circ} \sin 45^{\circ} \\ &= \left(-\frac{1}{2} \right) \left(\frac{1}{\sqrt{2}} \right) + \left(\frac{\sqrt{3}}{2} \right) \left(\frac{1}{\sqrt{2}}\right) \\ &= -\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}\\ &= \frac{\sqrt{3}-1}{2\sqrt{2}}. \end{align*}$$

Use the value of $\cos 75^{\circ}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ to find $\cos 105^{\circ}$ by using $\cos \left(180^{\circ}-75^{\circ}\right)$.

Solution.

$$\begin{align*} \cos 105^\circ & = \cos(180^\circ - 75^\circ) \\ & = -\cos 75^\circ \quad (\because \cos(180^\circ - \theta) = -\cos \theta) \\ & = -\left( \dfrac{\sqrt{3} - 1}{2\sqrt{2}} \right) \\ & = \dfrac{-(\sqrt{3} - 1)}{2\sqrt{2}} \\ & = \dfrac{1 - \sqrt{3}}{2\sqrt{2}}. \end{align*}$$

Use the value of $\cos 75^{\circ}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ to find $\cos 285^{\circ}$ by using $\cos \left(360^{\circ}-75^{\circ}\right)$.

Solution.

$$\begin{align*} \cos 285^{\circ} & = \cos \left(360^{\circ} - 75^{\circ}\right) \\ & = \cos 75^{\circ} \quad (\because \cos(360^{\circ} - \theta) = \cos \theta) \\ & = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}. \end{align*}$$

Find the exact value of $\sin 15^{\circ}$.

Solution.

$$\begin{align*} \sin 15^{\circ} & = \sin \left(45^{\circ} - 30^{\circ}\right) \\ &= \sin 45 \cos 30 - \cos 45 \sin 30 \\ &= \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \\ &= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \\ &= \frac{\sqrt{3} - 1}{2\sqrt{2}}. \end{align*}$$

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