====== 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. 

===== Question 3(a)=====
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*}
GOOD
===== Question 3(b)=====
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*}

===== Question 3(c)=====
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*}
GOOD
===== Question 3(d)=====
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*}

===== Question 3(e)=====
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*}

===== Question 3(f)=====
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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