Exercise (11.3) - No(1) Sample Solutions


(b) Solution (1)

$ \displaystyle \sin \text{135}{}^\circ =\sin (\text{180}{}^\circ -45{}^\circ )=\sin 45{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \cos 135{}^\circ \text{ =}\cos \text{(180}{}^\circ -45{}^\circ )=-\cos 45{}^\circ =-\frac{{\sqrt{2}}}{2}$

$ \displaystyle \begin{array}{l}\tan 135{}^\circ \text{ =}-1\\\\\cot 135{}^\circ \text{ =}-1\\\\\sec 135{}^\circ \text{ =}-\sqrt{2}\\\\\operatorname{cosec}135{}^\circ \text{ =}\sqrt{2}\end{array}$

(b) Solution (2)

$ \displaystyle \begin{array}{l}\ \ \ \text{principal angle = 135}{}^\circ \\\\\therefore \text{basic acute angle = 45}{}^\circ \end{array}$

$ \displaystyle \therefore \ \sin \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}\operatorname{and}\ \cos \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \begin{array}{l}\ \ \ \sin \text{135}{}^\circ =\sin \text{45}{}^\circ \operatorname{and}\ \cos \text{135}{}^\circ =\cos \text{45}{}^\circ \ \text{numerically}\text{.}\\\\\ \ \ \text{But 135}{}^\circ \ \text{lies in the second quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \sin \text{135}{}^\circ =\sin \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \cos \text{135}{}^\circ =-\cos \text{45}{}^\circ =-\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \begin{array}{*{20}{l}} {\tan {{{135}}^{{}^\circ }}\text{ =}-1} \\ {} \\ {\cot {{{135}}^{{}^\circ }}\text{ =}-1} \\ {} \\ {\sec {{{135}}^{{}^\circ }}\text{ =}-\sqrt{2}} \\ {} \\ {\operatorname{cosec}{{{135}}^{{}^\circ }}\text{ =}\sqrt{2}} \end{array}$

(d) Solution (1)

$ \displaystyle \ \ \ \sin 210{}^\circ =\sin (180{}^\circ +30{}^\circ )=-\sin 30{}^\circ =\displaystyle -\frac{1}{2}$

$ \displaystyle \ \ \ \cos 210{}^\circ =\cos (180{}^\circ +30{}^\circ )=-\cos 30{}^\circ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \tan 210{}^\circ =\displaystyle \frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \cot 210{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \sec 210{}^\circ =\displaystyle -\frac{{2\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \operatorname{cosec} 210{}^\circ =-2$

(d) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = 210}{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 30}{}^\circ $

$ \displaystyle \therefore \ \sin \text{30}{}^\circ =\frac{1}{2}\operatorname{and}\ \cos \text{30}{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \sin \text{210}{}^\circ =\sin \text{30}{}^\circ \operatorname{and}\ \cos \text{210}{}^\circ =\cos \text{30}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But 210}{}^\circ \ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore \sin \text{210}{}^\circ =\sin \text{30}{}^\circ =-\frac{1}{2}$

$ \displaystyle \ \ \ \cos \text{210}{}^\circ =-\cos \text{30}{}^\circ =\displaystyle -\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \begin{array}{*{20}{l}} {\tan \text{210}{}^\circ \text{ =}\displaystyle \frac{{\sqrt{3}}}{3}} \\ {} \\ {\cot \text{210}{}^\circ \text{ =}\sqrt{3}} \\ {} \\ {\sec \text{210}{}^\circ \text{ =}\displaystyle -\frac{{2\sqrt{3}}}{3}} \\ {} \\ {\operatorname{cosec}\text{210}{}^\circ \text{ =}-2} \end{array}$

(m) Solution (1)

$ \displaystyle \ \ \ \sin (-120{}^\circ )=-\sin 120{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin (180{}^\circ -60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \ \cos (-120{}^\circ )=\cos 120{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos (180{}^\circ -60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{2}$

$ \displaystyle \ \ \ \tan (-120{}^\circ )=\sqrt{3}$

$ \displaystyle \ \ \ \cot (-120{}^\circ )=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec (-120{}^\circ )=-2$

$ \displaystyle \ \ \ \operatorname{cosec}(-120{}^\circ )=-\frac{{2\sqrt{3}}}{3}$

(m) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = }-120{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 60}{}^\circ $

$ \displaystyle \therefore \ \sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}\operatorname{and}\ \cos \text{60}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \sin (-120{}^\circ )=\sin \text{60}{}^\circ \operatorname{and}\ \cos (-120{}^\circ)x =\cos \text{60}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }(-120{}^\circ )\ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore \sin (-120{}^\circ )=\sin \text{60}{}^\circ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos (-120{}^\circ )=-\cos \text{60}{}^\circ =-\frac{1}{2}$

$ \displaystyle \ \ \ \tan (-120{}^\circ )=\sqrt{3}$

$ \displaystyle \ \ \ \cot (-120{}^\circ )=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec (-120{}^\circ )=-2$

$ \displaystyle \ \ \ \operatorname{cosec}(-120{}^\circ )=-\frac{{2\sqrt{3}}}{3}$

(q) Solution (1)

$ \displaystyle \ \ \ \sin 480{}^\circ =\sin (360{}^\circ +60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \ \cos 480{}^\circ =\cos (360{}^\circ +60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}$

$ \displaystyle \ \ \ \tan 480{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \cot 480{}^\circ =\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 480{}^\circ =2$

$ \displaystyle \ \ \ \operatorname{cosec}480{}^\circ =\frac{{2\sqrt{3}}}{3}$

(q) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = }480{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 60}{}^\circ $

$ \displaystyle \therefore \ \sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}\operatorname{and}\ \cos \text{60}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \sin 480{}^\circ =\sin \text{60}{}^\circ \operatorname{and}\ \cos 480{}^\circ =\cos \text{60}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But}\ 480{}^\circ \ \text{lies in the first quadrant}\text{.}$

$ \displaystyle \therefore \sin 480{}^\circ =\sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos 480{}^\circ =\cos \text{30}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \tan 480{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \cot 480{}^\circ =\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 480{}^\circ =2$

$ \displaystyle \ \ \ \operatorname{cosec}480{}^\circ =\frac{{2\sqrt{3}}}{3}$