Trigonometric Functions
In previous classes we have studied about the trigonometric ratio's in which we have studied about the various ratios of the sides of the triangle. In this chapter we will extend our studies till the relation between the various trigonometric ratios which is called trigonometric function and we will measure the angles in terms of radians.
| FUNTION | 1st | 2nd | 3rd | 4th |
| \[\sin \,\theta \] | + | + | - | - |
| \[\cos \,\theta \] | + | - | - | + |
| \[\tan \,\theta \] | + | - | + | - |
| \[\cos ec\,\theta \] | + | + | - | - |
| \[sec\,\theta \] | + | - | - | + |
| \[\cot \,\theta \] | + | + | - | - |
Domain and Range of Trigonometric Function
| FUNCTION | DOMAIN | RANGE |
| \[1.\,Sin\theta \] | R | \[\left[ -1,\,1 \right]\] |
| \[2.\,Cos\theta \] | R | \[\left[ -1,\,1 \right]\] |
| \[3.\,Tan\theta \] | \[R-\left\{ 0,\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},--- \right\}\] | \[(-\propto ,\propto )\] |
| \[4.Co\sec \theta \] | \[R-\left\{ o,\pi 2\pi ,3\pi ,--- \right\}\] | \[(-\propto ,-1][1,\propto )\] |
| \[5.Sec\theta \] | \[R-\left\{ o,\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},--- \right\}\] | \[(-\propto ,-1][1,\propto )\] |
| \[6.Cot\theta \] | \[R-\{0,\pi ,2\pi ,3\pi ,---\}\] | \[(-\propto ,\propto )\] |
Graphical Representation of 5m, Cos and Tan Function
1. \[Sin\theta \]
2. \[Cos\theta \]
Conversion of Trigonometric Function of Complementary and Supple- mentary Angles
(A) \[\theta =\left( \frac{\pi }{2}-\theta \right)\]
1. \[Sin\left( \frac{\pi }{2}-\theta \right)=\cos \theta \]
2. \[Cos\left( \frac{\pi }{2}-\theta \right)=\sin \theta \]
3. \[Tan\left( \frac{\pi }{2}-\theta \right)=Cot\theta \]
4. \[Cot\left( \frac{\pi }{2}-\theta \right)=Tan\theta \]
5. \[Sec\left( \frac{\pi }{2}-\theta \right)=Co\sec \theta \]
6. \[Co\sec \left( \frac{\pi }{2}+\theta \right)=\sec \theta \]
(B) \[\theta =\left( \frac{\pi }{2}+\theta \right)\]
1. \[Sin\left( \frac{\pi }{2}+\theta \right)=Cos\theta \]
2. \[Cos\left( \frac{\pi }{2}+\theta \right)=-Sin\theta \]
2. \[Tan\left( \frac{\pi }{2}+\theta \right)=-Cot\theta \]
4. \[Tan\left( \frac{\pi }{2}+\theta \right)=-Cot\theta \]
3. \[Sec\left( \frac{\pi }{2}+\theta \right)=-Co\sec \theta \]
5. \[Co\sec =\sec \theta \]
(C) \[\theta =\left( \pi -\theta \right)\]
1. \[Sin\,(\pi -\theta )=\sin \theta \]
2. \[Cos\,(\pi -\theta )=-Cos\theta \]
3. \[Tan\,(\pi -\theta )=-Tan\theta \]
4. \[Cot\,(\pi -\theta )=-Cot\theta \]
5. \[Sec\,(\pi -\theta )=-\sec \theta \]
6. \[Cosec\,(\pi -\theta )=-Co\sec \theta \]
(D) \[\theta =(\pi +\theta )\]
1. \[Sin\,(\pi +\theta )=-\sin \theta \]
2. \[Cos\,(\pi +\theta )=-Cos\theta \]
3. \[Tan\,(\pi +\theta )=-\tan \theta \]
4. \[Cot\,(\pi +\theta )=-\cot \theta \]
5. \[Sec\,(\pi +\theta )=-\sec \theta \]
6. \[Co\sec \,(\pi +\theta )=-Co\sec \theta \]
(E) \[\theta =\left( \frac{3\pi }{2}-\theta \right)\]
1. \[Sin=\left( \frac{3\pi }{2}-\theta \right)=-\cos \theta \]
2. \[Cos=\left( \frac{3\pi }{2}-\theta \right)=-\sin \theta \]
3. \[Tan=\left( \frac{3\pi }{2}-\theta \right)=\cot \theta \]
4. \[Cot=\left( \frac{3\pi }{2}-\theta \right)=\tan \theta \]
5. \[Sec=\left( \frac{3\pi }{2}-\theta \right)=-\cos ec\theta \]
6. \[Cosec=\left( \frac{3\pi }{2}-\theta \right)=-sec\theta \]
(F) \[\theta =\left( \frac{3\pi }{2}-\theta \right)\]
1. \[Sin=\left( \frac{3\pi }{2}+\theta \right)=-\cos \theta \]
2. \[Cos=\left( \frac{3\pi }{2}+\theta \right)=\sin \theta \]
3. \[Tan=\left( \frac{3\pi }{2}+\theta \right)=-\cot \theta \]
4. \[Cot=\left( \frac{3\pi }{2}+\theta \right)=-\tan \theta \]
5. \[Sec=\left( \frac{3\pi }{2}+\theta \right)=\cos ec\theta \]
6. \[Cosec=\left( \frac{3\pi }{2}+\theta \right)=-sec\theta \]
(G) \[\theta =(2\pi -\theta )\]
1. \[Sin(2\pi -\theta )=-\sin \theta \]
2. \[Cos(2\pi -\theta )=Cos\theta \]
3. \[Tan(2\pi -\theta )=-Tan\theta \]
4. \[Cot(2\pi -\theta )=-Cot\theta \]
5. \[Sec(2\pi -\theta )=Sec\theta \]
6. \[Cosec(2\pi -\theta )=Cosec\theta \]
1. \[Sin(A+B)=SinA.\,CosB+CosA.\,SinB\]
2. \[Sin(A-B)=SinA.\,CosB-CosA.\,SinB\]
3. \[Cos(A+B)=CosA.\,CosB-SinA.\,SinB\]
4. \[Cos(A-B)=CosA.\,CosB-SinA.\,SinB\]
5. \[Tan(A+B)=\frac{TanA+TanB}{1-TanA.TanB}\]
6. \[Tan(A-B)=\frac{TanA-TanB}{1+TanA.TanB}\]
7. \[Cot(A-B)=\frac{CotA.CotB-1}{CotA+CotB}\]
8. \[Cot(A+B)=\frac{CotA.CotB+1}{CotA-CotB}\]
9. \[Sin2A=SinA.CosA=\frac{2TanA}{1+Ta{{n}^{2}}A}\]
10. \[Cos\,2A=Co{{s}^{2}}A-Si{{n}^{2}}A=2Co{{s}^{2}}A-1\]\[=1-2{{\sin }^{2}}A=\frac{1-Ta{{n}^{2}}A}{1+Ta{{n}^{2}}A}\]
11. \[Tan2A=\frac{2TanA}{1-Ta{{n}^{2A}}}\]
12. \[Sin3A=3SinA-4Si{{n}^{3}}A\]
13. \[Cos3A=4Co{{s}^{3}}A-3CosA\]
14. \[Tan3A=\frac{3TanA-Ta{{n}^{3}}A}{1-3Ta{{n}^{2}}A}\]
(A) 1. \[SinA+SinB=2Sin\frac{A+B}{2}Cos\frac{A-B}{2}\]
2. \[SinA+SinB=2Sin\frac{A-B}{2}Cos\frac{A+B}{2}\]
3. \[CosA+CosB=2Cos\frac{A-B}{2}Cos\frac{A+B}{2}\]
4. \[CosA-CosB=-2\sin \frac{A-B}{2}Sin\frac{A+B}{2}\]
(B) 1. \[SinA.SinB=\frac{1}{2}[Cos(A-B)-Cos(A+B)]\]
2. \[CosA.CosB=\frac{1}{2}[Cos(A-B)+Cos(A+B)]\]
3. \[SinA.CosB=\frac{1}{2}[Sin(A-B)+Sin(A+B)]\]
(C) 1. \[Sin(-A)=-SinA\]
2. \[Cos(-A)=CosA\]
3. \[Tan(-A)=-TanA\]
(a) \[Tan\theta \]
(b) \[Sec\theta \]
(c) \[Sin\theta \]
(d) \[Cos\theta \]
(e) None of these
Answer: (a)
Explanation
We have, \[f\left( \theta \right)=\frac{\sin 5\theta -2\sin 3\theta +\sin \theta }{\cos 5\theta -\cos \theta }\]
\[\Rightarrow \,\,\,\frac{2\sin 3\theta \cos 2\theta -2\sin 3\theta }{-2\sin 3\theta \sin 2\theta }\]
\[\Rightarrow \,\,\,\frac{1-\cos 2\theta }{\sin 2\theta }=Tan\theta \]
(a) 1
(b) 5
(c) 10
(d) 8
(e) None of these
Answer: (c)
Explanation
We have, \[2{{\sin }^{2}}\frac{3\pi }{4}+2{{\cos }^{2}}\frac{\pi }{4}+2{{\sec }^{2}}\frac{\pi }{3}\]
\[=2{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+2{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+2{{\left( 2 \right)}^{2}}\]
= 10
(a) \[Cot3\theta \]
(b) \[Cos6\theta \]
(c) \[\sin 4\theta \]
(d) \[Tan3\theta \]
(e) None of these
Answer: (b)
Explanation
We have,
\[32{{\cos }^{6}}\theta -48{{\cos }^{4}}\theta +18{{\cos }^{2}}\theta -1\]
\[=32{{\cos }^{6}}\theta -4-48{{\cos }^{4}}\theta +24{{\cos }^{2}}\theta -6{{\cos }^{2}}\theta +3\]
\[=4(8co{{s}^{6}}\theta -1-12{{\cos }^{4}}\theta +6{{\cos }^{2}}\theta )-6{{\cos }^{2}}\theta +3\]
\[=4{{\left( 2{{\cos }^{2}}\theta -1 \right)}^{3}}-3\left( 2{{\cos }^{2}}\theta -1 \right)\]
\[=4{{\cos }^{3}}2\theta -3\cos 2\theta \]
\[=Cos6\theta \].
(a) 1
(b) \[-\frac{3}{2}\]
(c) \[\frac{3}{2}\]
(d) \[-\frac{1}{2}\]
(e) None of these
Answer: (c)
(a) 1
(b) 0
(c) – 1
(d) \[-\frac{1}{2}\]
(e) None of these
Answer: (b)
(a) \[n\pi +\frac{7\pi }{6}\]
(b) \[2n\pi -\frac{7\pi }{6}\]
(c) \[2n\pi +\frac{7\pi }{6}\]
(d) \[n\pi +{{\left( -1 \right)}^{n}}\frac{7\pi }{6}\]
(e) None of these
Answer: (d)
Explanation
We have, \[2{{\cos }^{2}}\theta -3Sin\theta =0\]
\[\Rightarrow \,\,2\left( 1-Si{{n}^{2}}\theta \right)-3Sin\theta =0\]
\[\Rightarrow \,\,\left( 2Sin\theta +1 \right)\left( \sin \theta -2 \right)=0\]
\[\Rightarrow \,\,\,\,\,\text{sin}\theta =-\frac{1}{2}\text{or}\,\sin \theta =2\]
But \[\sin \theta =2\] is not possible
Therefore \[Sin\theta =-\frac{1}{2}\]
\[\Rightarrow \,\,\,\theta =\frac{7\pi }{6}\]
Hence the general solution is \[n\pi +{{\left( -1 \right)}^{n}}\frac{7\pi }{6}\]
(a) \[n\pi \]
(b) \[-n\pi \]
(c) \[-n\pi +\frac{\pi }{6}\]
(d) \[-n\pi -\frac{\pi }{6}\]
(e) None of these
Answer: (a)
Explanation
We have, \[Tan\left( \frac{\pi }{4}+\theta \right)+Tan\left( \frac{\pi }{4}-\theta \right)=2\]
\[\Rightarrow \,\,\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=2\]
\[\Rightarrow \,\,\,\,\frac{{{\left( 1+\tan \theta \right)}^{2}}+{{\left( 1-\tan \theta \right)}^{2}}}{1-{{\tan }^{2}}\theta }=2\]
\[\Rightarrow \,\,\,\,1+{{\tan }^{2}}\theta =1-{{\tan }^{2}}\theta \]
\[\Rightarrow \,\,\,\tan \theta =0\]
\[\Rightarrow \,\,\,\theta =n\pi \]for all integer n
(a) \[\left( n\pi ,\pm \frac{\pi }{6} \right)\]
(b) \[\left( 2n\pi ,\pm \frac{\pi }{6},\frac{\pi }{4} \right)\]
(c) \[\left( -n\pi ,+\frac{\pi }{6},\frac{\pi }{2} \right)\]
(d) \[\left( n\pi ,\pm \frac{\pi }{6},\frac{n\pi }{4} \right)\]
(e) None of these
Answer: (d)
Explanation
We have \[Sin2\theta -Sin4\theta +Sin6\theta =0\]
\[\Rightarrow \,\,\,Sin4\theta \left( 2\cos 2\theta -1 \right)=0\]
\[\Rightarrow \,\,\theta =\frac{n\pi }{4}\,and\,\theta =n\pi \pm \frac{\pi }{6}\]
(a) \[\alpha =\beta \]
(b) \[\alpha =-\beta \]
(c) \[\alpha =-2\beta \]
(d) \[\alpha +\beta =\pi \]
(e) None of these
Answer: (a)
(a) \[\pi \]
(b) \[2\pi \] s
(c) \[\frac{\pi }{3}\]
(d) \[\frac{2\pi }{3}\]
(e) None of these
Answer: (c)

- Trigonometry was invented for the purposes of astronomy.
- The origins of trigonometry can be traced to the civilizations of ancient Egypt, Mesopotamia and the Indus valley more than 4000 years ago.
- The common practice of measuring angles in degrees, minutes and seconds comes from the Babylonian's base sixty system of numeration.
- The angle can be measured in radian.
- \[1\,radian=\left( \frac{{{180}^{o}}}{\pi } \right)and{{1}^{o}}=\frac{\pi }{180}radians\]
- \[Co{{s}^{2}}\theta +Si{{n}^{2}}\theta =1\]
- \[1+Ta{{n}^{2}}\theta =Se{{c}^{2}}\theta \]
- \[1+Co{{t}^{2}}\theta =Cose{{c}^{2}}\theta \]
(A) \[\theta =\left( \frac{\pi }{2}-\theta \right)\]
- \[Sin\left( \frac{\pi }{2}-\theta \right)=Cos\theta \]
- \[Cos\left( \frac{\pi }{2}-\theta \right)=Sin\theta \]
- \[Tan\left( \frac{\pi }{2}-\theta \right)=Cot\theta \]
- \[Cot\left( \frac{\pi }{2}-\theta \right)=Tan\theta \]
- \[Sec\left( \frac{\pi }{2}-\theta \right)=Co\sec \theta \]
- \[Cosec\left( \frac{\pi }{2}-\theta \right)=\sec \theta \]
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