Question 2, Exercise 10.1

Solutions of Question 2 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Evaluate exactly: $\sin \dfrac{\pi }{12}$

We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \frac{\pi }{3}\sin \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}\\ & =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align}

Evaluate exactly:$\tan {{75}^{\circ }}$

We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }}+{{30}^{\circ }}$ and using the identity:

\begin{align}\tan (\alpha +\beta )&=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }.\\ \tan (45+30)&=\dfrac{\tan 45+\tan 30}{1-\tan 45\tan 30}.\\ \Rightarrow \,\,\tan (75)&=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-\left( 1 \right)\left( \dfrac{1}{\sqrt{3}} \right)}\\ &=\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{3}}}{\dfrac{\sqrt{3}-1}{\sqrt{3}}}\\ &=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\\ &=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}\\ &={{\dfrac{\left( \sqrt{3}+1 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}}}^{2}}\\ &=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+2\left( 1 \right)\left( \sqrt{3} \right)+{{\left( 1 \right)}^{2}}}{3-1}\\ &=\dfrac{3+2\left( \sqrt{3} \right)+1}{2}\\ &=\dfrac{4+2\left( \sqrt{3} \right)}{2}\\ &=2+\left( \sqrt{3} \right)\end{align}

Evaluate exactly:$\tan {{105}^{\circ }}$

We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^\circ }}+{{45}^{{}^\circ }}$ and using the identity: \begin{align}\tan ({{60}^{\circ }}+{{45}^{\circ }})&=\dfrac{\tan {{60}^{\circ }}+\tan {{45}^{\circ }}}{1-\tan {{60}^{\circ }}\tan {{45}^{\circ }}}\\ \tan ({{105}^{\circ }})&=\dfrac{\sqrt{3}+1}{1-\left( \sqrt{3} \right)\left( 1 \right)}\\ &=\dfrac{\sqrt{3}+1}{1-\sqrt{3}}\\ &=\dfrac{\sqrt{3}+1}{1-\sqrt{3}}\times \dfrac{1+\sqrt{3}}{1+\sqrt{3}}\\ &=\dfrac{{{\left( 1+\sqrt{3} \right)}^{2}}}{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}\\ &=\dfrac{{{1}^{2}}+2\left( 1 \right)\left( \sqrt{3} \right)+{{\left( \sqrt{3} \right)}^{2}}}{1-3}\\ &=\dfrac{1+2\sqrt{3}+3}{-2}\\ &=\dfrac{4+\sqrt{3}}{-2}\\ &=-2-\sqrt{3}\end{align}

Evaluate exactly:$\tan \dfrac{5\pi }{12}$

We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left( 2+3 \right)\pi }{12}$ or $\dfrac{\pi }{6}+\dfrac{\pi }{4}$ and using the identity: \begin{align}\tan (\dfrac{\pi }{6}+\dfrac{\pi }{4})&=\dfrac{\tan \dfrac{\pi }{6}+\tan \dfrac{\pi }{4}}{1-\tan \dfrac{\pi }{6}\tan \dfrac{\pi }{4}}\\ \tan \dfrac{5\pi }{12}&=\dfrac{\dfrac{1}{\sqrt{3}}+1}{1-\left( \dfrac{1}{\sqrt{3}} \right)\left( 1 \right)}\\ &=\dfrac{\dfrac{1}{\sqrt{3}}+1}{1-\dfrac{1}{\sqrt{3}}}\\ &=\dfrac{\dfrac{1+\sqrt{3}}{\sqrt{3}}}{\dfrac{\sqrt{3}-1}{\sqrt{3}}}\\ &=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\\ &=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}\\ &=\dfrac{{{\left( \sqrt{3}+1 \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}}}\\ &=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+2\left( \sqrt{3} \right)\left( 1 \right)+{{\left( 1 \right)}^{2}}}{3-1}\\ &=\dfrac{3+2\sqrt{3}+1}{2}\\ &=\dfrac{4+2\sqrt{3}}{2}\\ &=2+\sqrt{3}\end{align}

Evaluate exactly: $\cos {{15}^{\circ }}$

We rewrite ${{15}^{\circ }}$ as $\left( {{60}^{\circ }}-{{45}^{\circ }} \right)$ and using the identity: \begin{align}\cos \left( {{60}^{\circ }}-{{45}^{\circ }} \right)&=\cos {{60}^{\circ }}\cos {{45}^{\circ }}+\sin {{60}^{\circ }}\sin {{45}^{\circ }}\\ \cos {{15}^{\circ }}&=\left( \dfrac{1}{2} \right)\left( \dfrac{1}{\sqrt{2}} \right)+\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{\sqrt{2}} \right)\\ &=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}\\ &=\dfrac{1+\sqrt{3}}{2\sqrt{2}}\\ &=\dfrac{\sqrt{2}\left( 1+\sqrt{3} \right)}{\sqrt{2}\left( 2\sqrt{2} \right)}\\ &=\dfrac{\sqrt{2}+\sqrt{6}}{4}\end{align}

Evaluate exactly: $\sin \dfrac{7\pi }{12}$

We rewrite $\dfrac{7\pi }{12}$ as $\left( \dfrac{\pi }{4}+\dfrac{\pi }{3} \right)$ and using the identity: \begin{align}\sin (\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \sin \left( \dfrac{\pi }{4}+\dfrac{\pi }{3} \right)&=\sin \dfrac{\pi }{4}\cos \dfrac{\pi }{3}+\cos \dfrac{\pi }{4}\sin \dfrac{\pi }{3}\\ \sin \dfrac{7\pi }{12}&=\dfrac{1}{\sqrt{2}}\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}\\ &=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}\\ &=\dfrac{\sqrt{2}\left( 1+\sqrt{3} \right)}{\sqrt{2}\left( 2\sqrt{2} \right)}\\ &=\dfrac{\sqrt{2}+\sqrt{6}}{4}\end{align}

< Question 1 Question 3 >