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Chapter 3 Trigonometric Functions Ex 3.3 Solutions

Question - 21 : - Prove that: 

Answer - 21 : -


Question - 22 : - Prove that: cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Answer - 22 : -


Question - 23 : - Prove that: 

Answer - 23 : -

Consider
LHS = tan 4x = tan 2(2x)
By using the formula

Question - 24 : - Prove that: cos 4x = 1 – 8sin2 x cos2 x

Answer - 24 : -

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos2A = 1 – 2 sin2 A

= 1 – 2 sin2 2x

Again by using theformula sin2A = 2sin A cos A

= 1 – 2(2 sin x cos x) 2

So we get

= 1 – 8 sin2x cos2x

= R.H.S.

Question - 25 : - Prove that: cos 6x = 32 cos6 x – 48 cos4 x +18 cos2 x – 1

Answer - 25 : -

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos3A = 4 cos3 A – 3 cos A

= 4 cos3 2x –3 cos 2x

Again by using formulacos 2x = 2 cos2 x – 1

= 4 [(2 cos2 x –1)3 – 3 (2 cos2 x – 1)

By furthersimplification

= 4 [(2 cos2 x) 3 –(1)3 – 3 (2 cos2 x) 2 +3 (2 cos2 x)] – 6cos2 x + 3

We get

= 4 [8cos6x –1 – 12 cos4x + 6 cos2x] – 6 cos2x +3

By multiplication

= 32 cos6x –4 – 48 cos4x + 24 cos2 x – 6cos2x + 3

On further calculation

= 32 cos6x –48 cos4x + 18 cos2x – 1

= R.H.S.

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