Question 6, Exercise 10.2

Solutions of Question 6 of Exercise 10.2 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.

Use the half angle identities to evaluate exactly $\cos {{15}^{\circ }}$.

Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ }}}{2}$, and $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, $$\begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\circ }}}{2}=\sqrt{\dfrac{1+\cos {{30}^{\circ }}}{2}}\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\end{align}$$

Use the half angle identities to evaluate exactly $\tan {{67.5}^{\circ }}$.

Because ${{67.5}^{\circ }}=\dfrac{{{135}^{\circ }}}{2}$, the $\dfrac{\theta }{2}=\dfrac{{{135}^{\circ }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half angle identity as, $$\begin{align}\tan {{67.5}^{\circ }}&=\tan \dfrac{{{135}^{\circ }}}{2}=\sqrt{\dfrac{1-\cos {{135}^{\circ }}}{1+\cos {{135}^{\circ }}}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{1+\left( -\dfrac{1}{\sqrt{2}} \right)}}=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{1-\dfrac{1}{\sqrt{2}}}}\\ &=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}}\\ &=\sqrt{\dfrac{2+1+2\sqrt{2}}{2-1}}=\sqrt{3+2\sqrt{2}}\end{align}$$

Use the half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$.

Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfrac{\theta }{2}=\dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, $$\begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\sqrt{\dfrac{1-\cos {{225}^{\circ }}}{2}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{2}}=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\\ &=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}\end{align}$$

Use the half angle identities to evaluate exactly $\cos \dfrac{\pi }{8}$.

Because $\dfrac{\pi }{8}=\dfrac{\dfrac{\pi }{4}}{2}$, the $\dfrac{\theta }{2}=\dfrac{\dfrac{\pi }{4}}{2}$, so we can find $\cos \dfrac{\pi }{8}$by using half angle identity as, $$\begin{align} \cos \dfrac{\pi }{8}&=\cos \dfrac{\dfrac{\pi }{4}}{2}=\sqrt{\dfrac{1+\cos \dfrac{\pi }{4}}{2}}\\ &=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}} =\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\\ &=\dfrac{\sqrt{2+\sqrt{2}}}{2} \end{align}$$

Use the half angle identities to evaluate exactly $\tan {{75}^{\circ }}$.

Because ${{75}^{\circ }}=\dfrac{{{150}^{\circ }}}{2}$, the $\dfrac{\theta }{2}=\dfrac{{{150}^{\circ }}}{2}$ lies in first quadrant, so we can find $\tan {{75}^{\circ }}$by using half angle identity as, $$\begin{align}\tan {{75}^{\circ }}&=\tan \dfrac{{{150}^{\circ }}}{2}\\ &=\sqrt{\dfrac{1-\cos {{150}^{\circ }}}{1+\cos {{150}^{\circ }}}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{1+\left( -\dfrac{\sqrt{3}}{2} \right)}}\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{1-\dfrac{\sqrt{3}}{2}}}\\ &=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=}\sqrt{7+2\sqrt{3}}\end{align}$$

Use the half angle identities to evaluate exactly $\sin \dfrac{5\pi }{12}$.

Because $\dfrac{5\pi }{12}=\dfrac{\dfrac{5\pi }{6}}{2}$, the $\dfrac{\theta }{2}=\dfrac{\dfrac{5\pi }{6}}{2}$, so we can find $\sin \dfrac{5\pi }{12}$by using half angle identity as, $$\begin{align} \sin \dfrac{5\pi }{12}&=\sin \dfrac{\dfrac{5\pi }{6}}{2}=\sqrt{\dfrac{1-\cos \dfrac{5\pi }{12}}{2}}\\&=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{2}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{2}}\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\end{align}$$

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