NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q1:  
Ans: L.H.S. =  


Q2: Prove that  
Ans: L.H.S. =  


Q3: Prove that  
Ans: L.H.S. =


Q4: Prove that  
Ans: L.H.S =


Q5: Find the value of:
(i) sin 75°  
(ii) tan 15°
Ans:  (i) sin 75° = sin (45°+ 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x  + y) = sin x cos y  + cos x sin y]

 
(ii) tan 15° = tan (45° – 30°)


Q6: Prove that:
Ans: 



Q7: Prove that:  
Ans: It is known that  
∴ L.H.S. =

Q8: Prove that
Ans: 


Q9: 
Ans: L.H.S. =  


Q10: Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Ans: L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x


Q11: Prove that
Ans: It is known that  .

∴L.H.S. =  

             

Q12: Prove that sin² 6x – sin² 4x = sin 2x sin 10x
Ans: It is known that 

 
∴ L.H.S. = sin²6x – sin²4x
= (sin 6x + sin 4x) (sin 6x – sin 4x)  
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.

Q13: Prove that cos² 2x – cos² 6x = sin 4x sin 8x
Ans: It is known that 

 
∴ L.H.S. = cos² 2x – cos² 6x
= (cos 2x + cos 6x) (cos 2x – 6x)

= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.

Q14: Prove that sin 2x + 2sin 4x + sin 6x = 4cos² x sin 4x
Ans: L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] +2 sin 4x

= 2 sin 4x cos (– 2x)+ 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos² x – 1+ 1)
= 2 sin 4x (2 cos² x)
= 4cos² x sin 4x
= R.H.S.

Q15: Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Ans: L.H.S = cot 4x (sin 5x  sin 3x)

= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)

= 2 cos 4x. cos x
L.H.S. = R.H.S.

Q16:  Prove that  
Ans: It is known that

∴ L.H.S =


Q17: Prove that 
Ans: It is known that

∴L.H.S. =

             

Q18: Prove that 
Ans: It is known that

∴ L.H.S. =

               

Q19: Prove that   
Ans: It is known that

∴L.H.S. =

             

Q20: Prove that  
Ans: It is known that

∴L.H.S. =  

             

Q21: Prove that  
Ans: L.H.S. =


Q22: Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Ans: L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2x  + x) (cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1)
= 1 = R.H.S.

Q23: Prove that  
Ans: It is known that  .
∴ L.H.S. = tan 4x = tan 2(2x)

             

Q24: Prove that cos 4x = 1 – 8sin² x cos² x
Ans: L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin² 2x [cos 2A = 1 – 2 sin² A]
= 1 – 2(2 sin x cos x)² [sin2A = 2sin A cosA]
= 1 – 8 sin²x cos²x
= R.H.S.

Q25: Prove that: cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1
Ans: L.H.S. = cos 6x
= cos 3(2x)
= 4 cos³ 2x – 3 cos 2x [cos 3A = 4 cos³ A – 3 cos A]
= 4 [(2 cos² x – 1)³ – 3 (2 cos² x – 1) [cos 2x = 2 cos² x – 1]
= 4 [(2 cos² x)³ – (1)³ – 3 (2 cos² x)² + 3 (2 cos² x)] – 6cos² x + 3
= 4 [8cos⁶x – 1 – 12 cos⁴x + 6 cos²x] – 6 cos²x + 3
= 32 cos⁶x – 4 – 48 cos⁴x + 24 cos² x – 6 cos²x + 3
= 32 cos⁶x – 48 cos⁴x + 18 cos²x – 1
= R.H.S.