Question

Find the values of the following trigonometric ratios:

$\left(i\right)sin\frac{5\pi }{3}$

$\left(ii\right)sin{3060}^{\circ }$

$\left(iii\right)tan\frac{11\pi }{6}$

$\left(iv\right)cos(-{1125}^{\circ })$

$\left(V\right)tan{315}^{\circ }$

$\left(vi\right)sin{510}^{\circ }$

$\left(vii\right)cos{570}^{\circ }$

$\left(viii\right)sin(-{330}^{\circ })$

$\left(ix\right)cosec(-{1200}^{\circ })$

$\left(X\right)tan(-{585}^{\circ })$

$\left(xi\right)cos{855}^{\circ }$

$\left(xii\right)sin{1845}^{\circ }$

$\left(xiii\right)cos{1755}^{\circ }$

$\left(xiv\right)sin{4530}^{\circ }$

Answer 1

$\left(i\right)sin\frac{5\pi }{3}=sin(2\pi -\frac{\pi }{3})$

$=-sin\frac{\pi }{3}[\because sin(2\pi -\theta )=-sin\theta ]$

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

$\left(ii\right){3060}^{\circ }=17\pi [\because \pi ={180}^{\circ }]$

$\therefore sin{3060}^{\circ }=sin17\pi$

=0

$[\because sinn\pi =0foralln\in z]$

$\left(iii\right)tan\frac{11\pi }{6}=tan(2\pi -\frac{\pi }{6})$

$=-tan\frac{\pi }{6}$

$[\because tan(2\pi -\theta )=-tan\theta ]$

$=\frac{-1}{\sqrt{3}}$

$\left(iv\right){1125}^{\circ }=6\pi +\frac{\pi }{4}(\pi ={180}^{\circ })$

$cos(-{1125}^{\circ })=cos(-(6\pi +\frac{\pi }{4}))$

$=cos(6\pi +\frac{\pi }{4})$

$[\because cos(-\theta )=cos\theta ]$

$=cos(2\times 3\pi +\frac{\pi }{4})$

$=cos\frac{\pi }{4}$

$[\because cos(2k\pi +\theta )=cos\theta ,k\in n]$

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

$\left(v\right)tan{315}^{\circ }=tan(2\pi -\frac{\pi }{4})$

$=-tan\frac{\pi }{4}$

$[\because tan(2\pi -\theta )=-tan\theta ]$

=-1

$\left(vi\right)sin{510}^{\circ }=sin(3\pi -\frac{\pi }{6})$

$=sin\frac{\left(\pi \right)}{6}$

$[\because 3\pi -\frac{\pi }{6}$ lies in second quadrant]

$=\frac{1}{2}$

Alternative solution

$sin{510}^{\circ }=sin(3\pi -\frac{\pi }{6})$

$=sin(2\pi +(\pi -\frac{\pi }{6}))$

$=sin(\pi -\frac{\pi }{6})$

$[\because sin(2\pi +\theta )=sin\theta$, as sine is periodic with period $2\pi ]$

$=sin\frac{\pi }{6}[\because sin(\pi -\theta )=sin\theta ]$

$=\frac{1}{2}$

$\left(vii\right)cos{570}^{\circ }=cos(3\pi +\frac{\pi }{6})$

$=cos(2\pi +(\pi +\frac{\pi }{6}))$

$=cos(\pi +\frac{\pi }{6})$

$[\because cos(2\pi +\theta )=cos\theta$, as cosine is periodoc with period $2\pi ]$

$=-cos\frac{\pi }{6}$

$[\because cos(\pi +\theta )=-cos\theta ]$

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

$\left(viii\right)sin(-{330}^{\circ })=sin(-(2\pi -\frac{\pi }{6}))$

$=sin(2\pi -\frac{\pi }{6})[\because sin(-\theta )=-sin\theta ]$

$=-(-sin\frac{\pi }{6})$

$[\because sin(2\pi -\theta )=-sin\theta ]$

$=sin\frac{\pi }{6}$

$=\frac{1}{2}$

$\left(ix\right)cosec(-{1200}^{\circ })=cosec(-(7\pi -\frac{\pi }{3}))$

$=cosec(7\pi -\frac{\pi }{3})$

$[\because cosec(-\theta )=-cosec\theta ]$

$=-cosec(2\times 3\pi +(\pi -\frac{\pi }{3}))$

$=-cosec(\pi -\frac{\pi }{3})$

$[\because$ cosec is periodic of period $2\pi ,\therefore cosec(2\pi +\theta )=cosec(2n\pi +\theta )$

$=cosec\theta foralln\in N]$

$=-cosec\frac{\pi }{3}$

$[\because cosec(\pi -\theta \left)cosec\theta \right]$

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

$\left(x\right)tan(-{585}^{\circ })=-tan\left(585\right)$

$[\because tan(-\theta )=-tan\theta ]$

$=-tan(3\pi +\frac{\pi }{4})$

$=-tan(2\pi +(\pi +\frac{\pi }{4}))$

$[\because tan(2\pi +\theta )=tan\theta ]$

$=-tan\frac{\pi }{4}$

$[\because tan(\pi +\theta )=tan\theta ]$

=-1

$\left(xi\right)cos\left({855}^{\circ }\right)=cos(5\pi -\frac{\pi }{4})$

$=cos(2\times 2\pi +(\pi -\frac{\pi }{4}))$

$=cos(\pi -\frac{\pi }{4})$

$[\because cos(2k\pi +\theta )=cos\theta$ for all $k\in N]$

$=-cos\frac{\pi }{4}$

$[\because cos(\pi -\theta )=-cos\theta ]$

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

$\left(xii\right)sin{1845}^{\circ }=sin(10\pi +\frac{\pi }{4})$

$=(2\pi 5\pi +\frac{\pi }{4})$

$=sin\pi$

$[\because sin(2k\pi +\theta )=sin\theta$, for all $k\in N]$

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

$\left(xiii\right)cos{1755}^{\circ }=cos(10\pi -\frac{\pi }{4})$

$=cos(2\pi 5\pi -\frac{\pi }{4})$

$=cos\frac{\pi }{4}$

$[\because cos(2k\pi -\theta )=cos\theta ,k\in N]$

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

$\left(xiv\right){4530}^{\circ }=(25\pi +\frac{\pi }{6})$

$\therefore sin4530=sin(25\pi +\frac{\pi }{6})$

$=sin2\times 12\pi +(\pi +\frac{\pi }{6})$

$=sin\left(\pi \frac{\pi }{6}\right)$

$[\because sin(2k\pi +\theta )=sin\theta ,k\in N]$

$=-sin\frac{\pi }{6}[\because sin(\pi +\theta )=-sin\theta ]$

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