10th Class Mathematics Trigonometric Function Trigonometric Functions

Trigonometric Functions

Category : 10th Class

*      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.  

 

*        Sign of Trigonometric Function in Different Quadrants  

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. \[Cosec(2\pi -\theta )=Cosec\theta \]

 

*        Trigonometric Functions of Sum and Difference of two Angles

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. \[Tan3A=\frac{3TanA-Ta{{n}^{3}}A}{1-3Ta{{n}^{2}}A}\]

 

*        Some More Relations on Functions

 

(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. \[Tan(-A)=-TanA\]

 

 

  

 

Find the value of \[f\left( \theta  \right)=\frac{\sin 5\theta -2\sin 3\theta +\sin \theta }{\cos 5\theta -\cos \theta }\].

(a) \[Tan\theta \]                                               

(b) \[Sec\theta \]

(c) \[Sin\theta \]                                                 

(d) \[Cos\theta \]

(e) None of these

 

Answer: (a)

Explanation

We have, \[f\left( \theta  \right)=\frac{\sin 5\theta -2\sin 3\theta +\sin \theta }{\cos 5\theta -\cos \theta }\]

\[\Rightarrow \,\,\,\frac{2\sin 3\theta \cos 2\theta -2\sin 3\theta }{-2\sin 3\theta \sin 2\theta }\]

\[\Rightarrow \,\,\,\frac{1-\cos 2\theta }{\sin 2\theta }=Tan\theta \]

 

 

The value of \[2{{\sin }^{2}}\frac{3\pi }{4}+2{{\cos }^{2}}\frac{\pi }{4}+2{{\sec }^{2}}\frac{\pi }{3}\] is given by:

(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

 

 

The value of \[32{{\cos }^{6}}\theta -48{{\cos }^{4}}\theta +18{{\cos }^{2}}\theta -1\] is

(a) \[Cot3\theta \]                             

(b) \[Cos6\theta \]

(c) \[\sin 4\theta \]                            

(d) \[Tan3\theta \]                            

(e) None of these

 

Answer: (b)

Explanation

We have, 

\[32{{\cos }^{6}}\theta -48{{\cos }^{4}}\theta +18{{\cos }^{2}}\theta -1\]

\[=32{{\cos }^{6}}\theta -4-48{{\cos }^{4}}\theta +24{{\cos }^{2}}\theta -6{{\cos }^{2}}\theta +3\]

\[=4(8co{{s}^{6}}\theta -1-12{{\cos }^{4}}\theta +6{{\cos }^{2}}\theta )-6{{\cos }^{2}}\theta +3\]

\[=4{{\left( 2{{\cos }^{2}}\theta -1 \right)}^{3}}-3\left( 2{{\cos }^{2}}\theta -1 \right)\]

\[=4{{\cos }^{3}}2\theta -3\cos 2\theta \]

\[=Cos6\theta \].

 

Find the value of \[Co{{s}^{2}}\theta +Co{{s}^{2}}\left( \theta +\frac{\pi }{3} \right)+Co{{s}^{2}}\left( \theta -\frac{\pi }{3} \right)\].

(a) 1                                         

(b) \[-\frac{3}{2}\]

(c) \[\frac{3}{2}\]                                                

(d) \[-\frac{1}{2}\]

(e) None of these

 

Answer: (c)

 

 

Find the value of  \[2Cos\frac{\pi }{13}Cos\frac{9\pi }{13}+Cos\frac{3\pi }{13}+Cos\frac{5\pi }{13}\].

(a) 1                                                         

(b) 0         

(c) – 1                                                      

(d) \[-\frac{1}{2}\]

(e) None of these

 

Answer: (b)

 

 

The general solution of \[2{{\cos }^{2}}\theta -3\sin \theta =0\] is given by:

(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 \[n\pi +{{\left( -1 \right)}^{n}}\frac{7\pi }{6}\]

 

 

 

If \[Tan\left( \frac{\pi }{4}+\theta  \right)+Tan\left( \frac{\pi }{4}-\theta  \right)=2\], then the general solution is given by:

(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 

 

 

 

Find the general solution of the equation\[Sin\theta -Sin4\theta +Sin6\theta =0\].

(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 \[Sin2\theta -Sin4\theta +Sin6\theta =0\]

\[\Rightarrow \,\,\,Sin4\theta \left( 2\cos 2\theta -1 \right)=0\]

\[\Rightarrow \,\,\theta =\frac{n\pi }{4}\,and\,\theta =n\pi \pm \frac{\pi }{6}\]

 

 

If \[Sin\alpha =Sin\beta \,\,and\,Cos\alpha =Cos\beta \], then which one of the following options is correct and satisfies the above equation.

(a) \[\alpha =\beta \]                                       

(b) \[\alpha =-\beta \]

(c) \[\alpha =-2\beta \]                                    

(d) \[\alpha +\beta =\pi \]

(e) None of these

Answer: (a)

 

 

If \[Cos\theta +\sqrt{s3}\sin \theta =2\,and\,\theta \in \left[ 0,2\pi  \right]\], then the value of 9 is given by:

(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)          \[\theta =\left( \frac{\pi }{2}-\theta  \right)\]

  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.  \[Cosec\left( \frac{\pi }{2}-\theta  \right)=\sec \theta \]

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