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NCERT solution class 11 chapter 3 Trigonometric Functions exercise 3.3 mathematics

EXERCISE 3.3


Page No 73:

Question 1:

Answer:

L.H.S. = 

Question 2:

Prove that 

Answer:

L.H.S. = 

Question 3:

Prove that 

Answer:

L.H.S. =

Question 4:

Prove that 

Answer:

L.H.S =

Question 5:

Find the value of:

(i) sin 75°

(ii) tan 15°

Answer:

(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°)

Question 6:

Prove that:

Answer:

Question 7:

Prove that: 

Answer:

It is known that 

∴L.H.S. =

Question 8:

Prove that

Answer:

Question 9:

Answer:

L.H.S. = 

Question 10:

Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Answer:

L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(+ 2)x

Question 11:

Prove that

Answer:

It is known that.

∴L.H.S. = 

Question 12:

Prove that sin2 6x – sin2 4x = sin 2x sin 10x

Answer:

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.

Question 13:

Prove that cos2 2x – cos2 6x = sin 4sin 8x

Answer:

It is known that 

∴L.H.S. = cos2 2x – cos2 6x

= (cos 2x + cos 6x) (cos 2– 6x)

= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= R.H.S.

Question 14:

Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Answer:

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 â€“ 1 + 1)

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

Question 15:

Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Answer:

L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x â€“ sin 3x)

= 2 cos 4x. cos x

L.H.S. = R.H.S.

Question 16:

Prove that 

Answer:

It is known that

∴L.H.S =

Question 17:

Prove that 

Answer:

It is known that

∴L.H.S. =

Question 18:

Prove that 

Answer:

It is known that

∴L.H.S. =

Question 19:

Prove that 

Answer:

It is known that

∴L.H.S. =

Question 20:

Prove that 

Answer:

It is known that

∴L.H.S. = 

Question 21:

Prove that 

Answer:

L.H.S. =

Page No 74:

Question 22:

Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Answer:

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) (cot 2x + cot x)

= cot x cot 2– (cot 2cot x – 1)

= 1 = R.H.S.

Question 23:

Prove that 

Answer:

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

Question 24:

Prove that cos 4x = 1 – 8sincosx

Answer:

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.

Question 25:

Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 – 1

Answer:

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 – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 – 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 cos6– 48 cos4x + 18 cos2x – 1

= R.H.S.


 

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