====== 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. 

===== Question 2(a)=====
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*}
GOOD
  
  
===== Question 2(b)=====
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*}
GOOD

**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*}
GOOD

===== Question 2(c)=====
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, $\cos 345^\circ = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$.
===== Question 2(d)=====
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*}
GOOD
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