Find the values of other five trigonometric functions.
1. cos x = $\displaystyle \small -\frac{1}{2}$ , x lies in third quadrant
We know that, $\displaystyle \small \sin^{2}x+\cos^{2}x=1$
⇒ sin x = $\displaystyle \small \pm \sqrt{1-\cos^{2} x}$
In third quadrant, sin x is negative. Therefore,
sin x = $\displaystyle \small - \sqrt{1-\cos^{2} x}$
= $\displaystyle \small - \sqrt{1-\left ( -\frac{1}{2} \right )^{2}}$ = $\displaystyle \small - \sqrt{1-\frac{1}{4}}$
= $\displaystyle \small - \sqrt{\frac{4-1}{4}}$ = $\displaystyle \small -\frac{\sqrt{3}}{2}$
sin x = $\displaystyle \small -\frac{\sqrt{3}}{2}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$ = $\displaystyle \small \frac{-\sqrt{3}/2}{-1/2}$
tan x = $\displaystyle \small \sqrt{3}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{-\sqrt{3}/2}$
cosec x = $\displaystyle \small -\frac{2}{\sqrt{3}}$
sec x = $\displaystyle \small \frac{1}{\cos x}$ = $\displaystyle \small \frac{1}{-1/2}$
sec x = -2
cot x = $\displaystyle \small \frac{1}{\tan x}$
cot x = $\displaystyle \small \frac{1}{\sqrt{3}}$
2. sin x = $\displaystyle \small \frac{3}{5}$ , x lies in second quadrant
We know that, $\displaystyle \small \sin^{2}x+\cos^{2}x=1$
⇒ cos x = $\displaystyle \small \pm \sqrt{1-\sin^{2} x}$
In second quadrant, cos x is negative. Therefore,
cos x = $\displaystyle \small - \sqrt{1-\sin^{2} x}$
= $\displaystyle \small - \sqrt{1-\left ( \frac{3}{5} \right )^{2}}$ = $\displaystyle \small - \sqrt{1-\frac{9}{25}}$
= $\displaystyle \small - \sqrt{\frac{25-9}{25}}$ = $\displaystyle \small -\sqrt{\frac{16}{25}}$
cos x = $\displaystyle \small -\frac{4}{5}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$ = $\displaystyle \small \frac{3/5}{-4/5}$
tan x = $\displaystyle \small -\frac{3}{4}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{3/5}$
cosec x = $\displaystyle \small \frac{5}{3}$
sec x = $\displaystyle \small \frac{1}{\cos x}$ = $\displaystyle \small \frac{1}{-4/5}$
sec x = $\displaystyle \small -\frac{5}{4}$
cot x = $\displaystyle \small \frac{1}{\tan x}$ = $\displaystyle \small \frac{1}{-3/4}$
cot x = $\displaystyle \small -\frac{4}{3}$
3. cot x = $\displaystyle \small \frac{3}{4}$ , x lies in third quadrant
We know that, $\displaystyle \small \csc^{2}x=1+\cot^{2}x$
⇒ cosec x = $\displaystyle \small \pm \sqrt{1+\cot^{2} x}$
In third quadrant, cosec x is negative. Therefore,
cosec x = $\displaystyle \small - \sqrt{1+\cot^{2} x}$
= $\displaystyle \small - \sqrt{1+\left ( \frac{3}{4} \right )^{2}}$ = $\displaystyle \small - \sqrt{1+\frac{9}{16}}$
= $\displaystyle \small - \sqrt{\frac{16+9}{16}}$ = $\displaystyle \small -\sqrt{\frac{25}{16}}$
cosec x = $\displaystyle \small -\frac{5}{4}$
sin x = 1/cosec x = $\displaystyle \small \frac{1}{-5/4}$
sin x = $\displaystyle \small -\frac{4}{5}$
tan x = $\displaystyle \small \frac{1}{\cot x}$ = $\displaystyle \small \frac{1}{3/4}$
tan x = $\displaystyle \small \frac{4}{3}$
Also, tan x = $\displaystyle \small \frac{\sin x}{\cos x}$
cos x = $\displaystyle \small \frac{\sin x}{\tan x}$ = $\displaystyle \small \frac{-4/5}{4/3}$
cos x = $\displaystyle \small -\frac{3}{5}$
sec x = $\displaystyle \small \frac{1}{\cos x}$ = $\displaystyle \small \frac{1}{-3/5}$
sec x = $\displaystyle \small -\frac{5}{3}$
4. sec x = $\displaystyle \small \frac{13}{5}$ , x lies in fourth quadrant
cos x = $\displaystyle \small \frac{1}{\sec x}$ = $\displaystyle \small \frac{1}{13/5}$
cos x = $\displaystyle \small \frac{5}{13}$
We know that, $\displaystyle \small \sin^{2}x+\cos^{2}x=1$
⇒ sin x = $\displaystyle \small \pm \sqrt{1-\cos^{2} x}$
In fourth quadrant, sin x is negative. Therefore,
sin x = $\displaystyle \small - \sqrt{1-\cos^{2} x}$
= $\displaystyle \small - \sqrt{1-\left ( \frac{5}{13} \right )^{2}}$ = $\displaystyle \small - \sqrt{1-\frac{25}{169}}$
= $\displaystyle \small - \sqrt{\frac{169-25}{169}}$ = $\displaystyle \small -\frac{\sqrt{144}}{169}$
sin x = $\displaystyle \small -\frac{12}{13}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$ = $\displaystyle \small \frac{-12/13}{5/13}$
tan x = $\displaystyle \small -\frac{12}{5}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{-12/13}$
cosec x = $\displaystyle \small -\frac{13}{12}$
cot x = $\displaystyle \small \frac{1}{\tan x}$ = $\displaystyle \small \frac{1}{-12/5}$
cot x = $\displaystyle \small -\frac{5}{12}$
5. tan x = $\displaystyle \small -\frac{5}{12}$ , x lies in second quadrant
cot x = $\displaystyle \small \frac{1}{\tan x}$ = $\displaystyle \small \frac{1}{-5/12}$
cot x = $\displaystyle \small -\frac{12}{5}$
We know that, $\displaystyle \small \sec^{2}x=1+\tan^{2}x$
⇒ sec x = $\displaystyle \small \pm \sqrt{1+\tan^{2} x}$
In second quadrant, sec x is negative. Therefore,
sec x = $\displaystyle \small - \sqrt{1+\tan^{2} x}$
= $\displaystyle \small - \sqrt{1+\left ( -\frac{5}{12} \right )^{2}}$ = $\displaystyle \small - \sqrt{1+\frac{25}{144}}$
= $\displaystyle \small - \sqrt{\frac{144+25}{144}}$ = $\displaystyle \small -\frac{\sqrt{169}}{144}$
sec x = $\displaystyle \small -\frac{13}{12}$
cos x = $\displaystyle \small \frac{1}{\sec x}$ = $\displaystyle \small \frac{1}{-13/12}$
cos x = $\displaystyle \small -\frac{12}{13}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$
⇒ sin x = tan x * cos x = $\displaystyle \small \frac{-5}{12}*\frac{-12}{13}$
sin x = $\displaystyle \small \frac{5}{13}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{5/13}$
cosec x = $\displaystyle \small \frac{13}{5}$
Find the value of the trigonometric functions.
6. sin 765°
We know that, the values of sin x repeats after a period of 2Ï€ or 360°
∴ sin 765° = sin (2*360°+45°)
= sin 45° = $\displaystyle \small \frac{1}{\sqrt{2}}$
7. cosec (-1410°)
cosec (-1410°) = -cosec (1410°) [∵ cosec (-x) = -cosec x]
We know that, the values of cosec x repeats after a period of 2Ï€ or 360°
-cosec (1410°) = -cosec (4*360°-30°)
= -cosec (-30°) = cosec (30°) = 2
8. tan $\displaystyle \small \left (\frac{19\pi }{3} \right )$
tan $\displaystyle \small \left (\frac{19\pi }{3} \right )$ = tan $\displaystyle \small \left ( 6\frac{\pi }{3} \right )$
= tan $\displaystyle \small \left ( 6\pi +\frac{\pi }{3} \right )$
We know that, the values of tan x repeats after a period of π
∴ tan $\displaystyle \small \left ( 6\pi +\frac{\pi }{3} \right )$ = tan $\displaystyle \small \left ( \frac{\pi }{3} \right )$
= $\displaystyle \small \sqrt{3}$
9. sin $\displaystyle \small \left (-\frac{11\pi }{3} \right )$
sin $\displaystyle \small \left (-\frac{11\pi }{3} \right )$ = - sin $\displaystyle \small \left (\frac{11\pi }{3} \right )$ [∵sin (-x) = -sin x]
= -sin $\displaystyle \small \left ( 4\frac{-\pi }{3} \right )$
= -sin $\displaystyle \small \left ( 4\pi -\frac{\pi }{3} \right )$
We know that, the values of sin x repeats after a period of 2Ï€
-sin $\displaystyle \small \left ( 4\pi -\frac{\pi }{3} \right )$ = -sin $\displaystyle \small \left ( -\frac{\pi }{3} \right )$
= sin $\displaystyle \small \left ( \frac{\pi }{3} \right )$ = $\displaystyle \small \frac{\sqrt{3}}{2}$
10. cot $\displaystyle \small \left (-\frac{15\pi }{4} \right )$
cot $\displaystyle \small \left (-\frac{15\pi }{4} \right )$ = - cot $\displaystyle \small \left (\frac{15\pi }{4} \right )$ [∵ cot (-x) = -cot x]
= -cot $\displaystyle \small \left ( 4\frac{-\pi }{4} \right )$
= -cot $\displaystyle \small \left ( 4\pi -\frac{\pi }{4} \right )$
We know that, the values of cot x repeats after a period of π
-cot $\displaystyle \small \left ( 4\pi -\frac{\pi }{4} \right )$ = -cot $\displaystyle \small \left ( -\frac{\pi }{4} \right )$
= cot $\displaystyle \small \left ( \frac{\pi }{4} \right )$ = 1
1. cos x = $\displaystyle \small -\frac{1}{2}$ , x lies in third quadrant
We know that, $\displaystyle \small \sin^{2}x+\cos^{2}x=1$
⇒ sin x = $\displaystyle \small \pm \sqrt{1-\cos^{2} x}$
In third quadrant, sin x is negative. Therefore,
sin x = $\displaystyle \small - \sqrt{1-\cos^{2} x}$
= $\displaystyle \small - \sqrt{1-\left ( -\frac{1}{2} \right )^{2}}$ = $\displaystyle \small - \sqrt{1-\frac{1}{4}}$
= $\displaystyle \small - \sqrt{\frac{4-1}{4}}$ = $\displaystyle \small -\frac{\sqrt{3}}{2}$
sin x = $\displaystyle \small -\frac{\sqrt{3}}{2}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$ = $\displaystyle \small \frac{-\sqrt{3}/2}{-1/2}$
tan x = $\displaystyle \small \sqrt{3}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{-\sqrt{3}/2}$
cosec x = $\displaystyle \small -\frac{2}{\sqrt{3}}$
sec x = $\displaystyle \small \frac{1}{\cos x}$ = $\displaystyle \small \frac{1}{-1/2}$
sec x = -2
cot x = $\displaystyle \small \frac{1}{\tan x}$
cot x = $\displaystyle \small \frac{1}{\sqrt{3}}$
2. sin x = $\displaystyle \small \frac{3}{5}$ , x lies in second quadrant
We know that, $\displaystyle \small \sin^{2}x+\cos^{2}x=1$
⇒ cos x = $\displaystyle \small \pm \sqrt{1-\sin^{2} x}$
In second quadrant, cos x is negative. Therefore,
cos x = $\displaystyle \small - \sqrt{1-\sin^{2} x}$
= $\displaystyle \small - \sqrt{1-\left ( \frac{3}{5} \right )^{2}}$ = $\displaystyle \small - \sqrt{1-\frac{9}{25}}$
= $\displaystyle \small - \sqrt{\frac{25-9}{25}}$ = $\displaystyle \small -\sqrt{\frac{16}{25}}$
cos x = $\displaystyle \small -\frac{4}{5}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$ = $\displaystyle \small \frac{3/5}{-4/5}$
tan x = $\displaystyle \small -\frac{3}{4}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{3/5}$
cosec x = $\displaystyle \small \frac{5}{3}$
sec x = $\displaystyle \small \frac{1}{\cos x}$ = $\displaystyle \small \frac{1}{-4/5}$
sec x = $\displaystyle \small -\frac{5}{4}$
cot x = $\displaystyle \small \frac{1}{\tan x}$ = $\displaystyle \small \frac{1}{-3/4}$
cot x = $\displaystyle \small -\frac{4}{3}$
3. cot x = $\displaystyle \small \frac{3}{4}$ , x lies in third quadrant
We know that, $\displaystyle \small \csc^{2}x=1+\cot^{2}x$
⇒ cosec x = $\displaystyle \small \pm \sqrt{1+\cot^{2} x}$
In third quadrant, cosec x is negative. Therefore,
cosec x = $\displaystyle \small - \sqrt{1+\cot^{2} x}$
= $\displaystyle \small - \sqrt{1+\left ( \frac{3}{4} \right )^{2}}$ = $\displaystyle \small - \sqrt{1+\frac{9}{16}}$
= $\displaystyle \small - \sqrt{\frac{16+9}{16}}$ = $\displaystyle \small -\sqrt{\frac{25}{16}}$
cosec x = $\displaystyle \small -\frac{5}{4}$
sin x = 1/cosec x = $\displaystyle \small \frac{1}{-5/4}$
sin x = $\displaystyle \small -\frac{4}{5}$
tan x = $\displaystyle \small \frac{1}{\cot x}$ = $\displaystyle \small \frac{1}{3/4}$
tan x = $\displaystyle \small \frac{4}{3}$
Also, tan x = $\displaystyle \small \frac{\sin x}{\cos x}$
cos x = $\displaystyle \small \frac{\sin x}{\tan x}$ = $\displaystyle \small \frac{-4/5}{4/3}$
cos x = $\displaystyle \small -\frac{3}{5}$
sec x = $\displaystyle \small \frac{1}{\cos x}$ = $\displaystyle \small \frac{1}{-3/5}$
sec x = $\displaystyle \small -\frac{5}{3}$
4. sec x = $\displaystyle \small \frac{13}{5}$ , x lies in fourth quadrant
cos x = $\displaystyle \small \frac{1}{\sec x}$ = $\displaystyle \small \frac{1}{13/5}$
cos x = $\displaystyle \small \frac{5}{13}$
We know that, $\displaystyle \small \sin^{2}x+\cos^{2}x=1$
⇒ sin x = $\displaystyle \small \pm \sqrt{1-\cos^{2} x}$
In fourth quadrant, sin x is negative. Therefore,
sin x = $\displaystyle \small - \sqrt{1-\cos^{2} x}$
= $\displaystyle \small - \sqrt{1-\left ( \frac{5}{13} \right )^{2}}$ = $\displaystyle \small - \sqrt{1-\frac{25}{169}}$
= $\displaystyle \small - \sqrt{\frac{169-25}{169}}$ = $\displaystyle \small -\frac{\sqrt{144}}{169}$
sin x = $\displaystyle \small -\frac{12}{13}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$ = $\displaystyle \small \frac{-12/13}{5/13}$
tan x = $\displaystyle \small -\frac{12}{5}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{-12/13}$
cosec x = $\displaystyle \small -\frac{13}{12}$
cot x = $\displaystyle \small \frac{1}{\tan x}$ = $\displaystyle \small \frac{1}{-12/5}$
cot x = $\displaystyle \small -\frac{5}{12}$
5. tan x = $\displaystyle \small -\frac{5}{12}$ , x lies in second quadrant
cot x = $\displaystyle \small \frac{1}{\tan x}$ = $\displaystyle \small \frac{1}{-5/12}$
cot x = $\displaystyle \small -\frac{12}{5}$
We know that, $\displaystyle \small \sec^{2}x=1+\tan^{2}x$
⇒ sec x = $\displaystyle \small \pm \sqrt{1+\tan^{2} x}$
In second quadrant, sec x is negative. Therefore,
sec x = $\displaystyle \small - \sqrt{1+\tan^{2} x}$
= $\displaystyle \small - \sqrt{1+\left ( -\frac{5}{12} \right )^{2}}$ = $\displaystyle \small - \sqrt{1+\frac{25}{144}}$
= $\displaystyle \small - \sqrt{\frac{144+25}{144}}$ = $\displaystyle \small -\frac{\sqrt{169}}{144}$
sec x = $\displaystyle \small -\frac{13}{12}$
cos x = $\displaystyle \small \frac{1}{\sec x}$ = $\displaystyle \small \frac{1}{-13/12}$
cos x = $\displaystyle \small -\frac{12}{13}$
tan x = $\displaystyle \small \frac{\sin x}{\cos x}$
⇒ sin x = tan x * cos x = $\displaystyle \small \frac{-5}{12}*\frac{-12}{13}$
sin x = $\displaystyle \small \frac{5}{13}$
cosec x = $\displaystyle \small \frac{1}{\sin x}$ = $\displaystyle \small \frac{1}{5/13}$
cosec x = $\displaystyle \small \frac{13}{5}$
Find the value of the trigonometric functions.
6. sin 765°
We know that, the values of sin x repeats after a period of 2Ï€ or 360°
∴ sin 765° = sin (2*360°+45°)
= sin 45° = $\displaystyle \small \frac{1}{\sqrt{2}}$
7. cosec (-1410°)
cosec (-1410°) = -cosec (1410°) [∵ cosec (-x) = -cosec x]
We know that, the values of cosec x repeats after a period of 2Ï€ or 360°
-cosec (1410°) = -cosec (4*360°-30°)
= -cosec (-30°) = cosec (30°) = 2
8. tan $\displaystyle \small \left (\frac{19\pi }{3} \right )$
tan $\displaystyle \small \left (\frac{19\pi }{3} \right )$ = tan $\displaystyle \small \left ( 6\frac{\pi }{3} \right )$
= tan $\displaystyle \small \left ( 6\pi +\frac{\pi }{3} \right )$
We know that, the values of tan x repeats after a period of π
∴ tan $\displaystyle \small \left ( 6\pi +\frac{\pi }{3} \right )$ = tan $\displaystyle \small \left ( \frac{\pi }{3} \right )$
= $\displaystyle \small \sqrt{3}$
9. sin $\displaystyle \small \left (-\frac{11\pi }{3} \right )$
sin $\displaystyle \small \left (-\frac{11\pi }{3} \right )$ = - sin $\displaystyle \small \left (\frac{11\pi }{3} \right )$ [∵sin (-x) = -sin x]
= -sin $\displaystyle \small \left ( 4\frac{-\pi }{3} \right )$
= -sin $\displaystyle \small \left ( 4\pi -\frac{\pi }{3} \right )$
We know that, the values of sin x repeats after a period of 2Ï€
-sin $\displaystyle \small \left ( 4\pi -\frac{\pi }{3} \right )$ = -sin $\displaystyle \small \left ( -\frac{\pi }{3} \right )$
= sin $\displaystyle \small \left ( \frac{\pi }{3} \right )$ = $\displaystyle \small \frac{\sqrt{3}}{2}$
10. cot $\displaystyle \small \left (-\frac{15\pi }{4} \right )$
cot $\displaystyle \small \left (-\frac{15\pi }{4} \right )$ = - cot $\displaystyle \small \left (\frac{15\pi }{4} \right )$ [∵ cot (-x) = -cot x]
= -cot $\displaystyle \small \left ( 4\frac{-\pi }{4} \right )$
= -cot $\displaystyle \small \left ( 4\pi -\frac{\pi }{4} \right )$
We know that, the values of cot x repeats after a period of π
-cot $\displaystyle \small \left ( 4\pi -\frac{\pi }{4} \right )$ = -cot $\displaystyle \small \left ( -\frac{\pi }{4} \right )$
= cot $\displaystyle \small \left ( \frac{\pi }{4} \right )$ = 1
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