Question 2, Exercise 8.1

Solutions of Question 2 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 $\cos 15^{\circ}$ by using $\cos \left(45^{\circ}-30^{\circ}\right)$.

Solution.

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

Use the value of $\cos 15^{\circ}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ to find $\cos 165^{\circ}$ by using $\cos \left(180^{\circ}-15^{\circ}\right)$.

Solution.

$$\begin{align*} \cos 165^{\circ} & = \cos \left(180^{\circ}-15^{\circ}\right)\\ &= -\cos 15 \quad (\because \cos(180-\theta)=-\cos\theta)\\ & = -\dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*}$$

Alternative Method (if $\cos 15^{\circ}$ is not given) $$\begin{align*} \cos 165^{\circ} & = \cos \left(180^{\circ}-15^{\circ}\right)\\ &= -\cos 15 \quad (\because \cos(180-\theta)=-\cos\theta)\\ & = -\left[\cos \left(45^{\circ}-30^{\circ}\right) \right]\\ &= -\left[\cos 45 \cos 30 + \sin 45 \sin 30 \right] \\ &= -\left[ \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{2} \right] \\ & = -\left[ \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \right] \\ & = -\dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*}$$

Use the value of $\cos 15^{\circ}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ to find $\cos 345^{\circ}$ by using $\cos \left(360^{\circ}-15^{\circ}\right)$.

Solution.

We are given that $\cos 15^\circ = \dfrac{\sqrt{3}+1}{2\sqrt{2}}$. Now, let's find $\cos 345^\circ$. $$\begin{align*} \cos 345^\circ & = \cos \left(360^\circ - 15^\circ\right) \\ & = \cos 15^\circ \quad (\because \cos(360^\circ - \theta) = \cos \theta) \\ & = \dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*}$$ Thus, $\cos 345^\circ = \dfrac{\sqrt{3}+1}{2\sqrt{2}}$.

Alternative Method (if $\cos 15^\circ$ is not given)

To find $\cos 345^\circ$, we use the identity $\cos(360^\circ - \theta) = \cos \theta$. $$\begin{align*} \cos 345^\circ & = \cos(360^\circ - 15^\circ) \\ & = \cos 15^\circ \\ & = \cos(45^\circ - 30^\circ) \\ & = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \\ & = \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} \\ & = \dfrac{\sqrt{3}}{2\sqrt{2}} + \dfrac{1}{2\sqrt{2}} \\ & = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}. \end{align*}$$ Thus, $\cos 345^\circ = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$.

Use $\cos A=\sin \left(90^{\circ}-A\right)$ to find the exact value of $\sin 75^{\circ}$ and then find $\tan 75^{\circ}$.

Solution.

Given $$\cos A=\sin \left(90^{\circ}-A\right) ... (1)$$ Put $A=15^\circ$, we get $$\begin{align*} & \sin \left(90-15\right) = \cos 15^\circ \\ \implies & \sin 75^\circ = \cos 15^\circ \\ & = \cos \left(45^{\circ}-30^{\circ}\right)\\ &= \cos 45 \cos 30 + \sin 45 \sin 30 \\ &= \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{2} \\ & = \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\\ \implies & \boxed{\sin 75^\circ = \dfrac{\sqrt{3}+1}{2\sqrt{2}}}. \end{align*}$$ Now put $A=75^\circ$ in (1) $$\begin{align*} \cos 75^\circ& =\sin \left(90-75\right) \\ & = \sin 15^\circ \\ & = \sin \left(45^{\circ}-30^{\circ}\right)\\ &= \sin 45 \cos 30 - \cos 45 \sin 30 \\ &= \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{2} \\ & = \dfrac{\sqrt{3}}{2\sqrt{2}}-\dfrac{1}{2\sqrt{2}}\\ & = \dfrac{\sqrt{3}-1}{2\sqrt{2}}. \end{align*}$$ Now $$\begin{align*} \tan 75^\circ & = \frac{\sin 75^\circ}{\cos 75^\circ} \\ & = \frac{(\sqrt{3}+1)/2\sqrt{2}}{(\sqrt{3}-1)/2\sqrt{2}} \end{align*}$$ $$\begin{align*} \implies \boxed{\tan 75^\circ = \frac{\sqrt{3}+1}{\sqrt{3}-1}} \end{align*}$$

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