We will learn how to
express the multiple angle of **cos 3A in
terms of A** or **cos 3A in terms of cos
A**.

Trigonometric function of cos 3A in terms of cos A is also known as one of the double angle formula.

If A is a number or angle then we have, cos 3A = 4 cos^3 A - 3 cos A

Now we will proof the above multiple angle formula step-by-step.

**Proof: **cos 3A

= cos (2A + A)

= cos 2A cos A - sin 2A sin A

= (2 cos^2 A - 1) cos A - 2 sin A cos A ∙ sin A

= 2 cos^3 A - cos A - 2 cos A (1 - cos^2 A)

= 2 cos^3 A - cos A - 2 cos A + 2 cos^3 A

= 4 cos^3 A - 3 cos A

Therefore, cos 3A = 4 cos^3 A - 3 cos A * Proved*

**Note:** (i) In
the above formula we should note that the angle on the R.H.S. of the
formula is one-third of the angle on L.H.S. Therefore, cos 120° = 4 cos^3 40° - 3 cos 40°.

(ii) To find the formula of cos 3A in terms of A or cos 3A in terms of cos A we have use cos 2A = 2cos^2 A - 1.

Now, we will apply the formula of multiple angle of cos 3A in terms of A or cos 3A in terms of cos A to solve the below problems.

**1.** Prove that: cos 6A = 32 cos^6 A - 48 cos^4 A + 18 cos^2 A
- 1

**Solution:**

L.H.S. = cos 6A

= 2 cos^2 3A - 1, [Since we know that, cos 2θ = 2 cos^2 θ - 1]

= 2(4 cos^3 A - 3 cos A)^2 - 1

= 2 (16 cos^ 6 A + 9 cos^2 A - 24 cos^2 A) - 1

= 32 cos^6 A – 48 cos^4 A + 18 cos^2 A - 1 = R.H.S.

**2.** Show that, 32
sin^6 θ = 10 - 15 cos 2θ + 6 cos 4θ - cos 6θ

**Solution: **

L.H.S = 32 sin^6 θ

= 4 ∙ (2 sin^2 θ)^3

= 4 (1 - cos 2θ)^3

= 4 [1 - 3 cos 2θ + 3 ∙ cos^2 2θ - cos^3 2θ]

= 4 - 12 cos^2 θ + 12 cos^2 2θ - 4 cos^3 2θ

= 4 - 12 cos 2θ + 6 ∙ 2 cos^2 2θ - [cos 3 ∙ (2θ) + 3 cos 2θ]

[Since, cos 3A = 4 cos^3 A - 3 cos A

Therefore, 4 cos^3 A = cos 3A + 3 cos A]

⇒ 4 cos^3 2θ = cos 3 ∙ (2θ) + 3 cos 2θ, (replacing A by 2θ)

= 4 - 12 cos 2θ + 6 (1 + cos 4θ) - cos 6θ - 3 cos 2θ

= 10 - 15 cos 2θ + 6 cos 4θ - cos 6θ = R.H.S. * Proved*

** **

**3. **Prove that: cos A cos (60 - A) cos (60 +
A) = ¼ cos 3A

**Solution:**

L.H.S. = cos A ∙ cos (60 - A) cos (60 + A)

= cos A ∙ (cos^2 60 - sin^2 A), [Since we know that cos (A + B) cos (A - B) = cos ^2 A - sin ^2 B]

= cos A (¼ - sin^2 A)

= cos A (¼ - (1 - cos^2 A))

= cos A (-3/4 + cos ^2 A)

= ¼ cos A (-3 + 4 cos^2 A)

= ¼(4 cos^3A - 3 cos A)

= ¼ cos 3A = R.H.S. * Proved*

**sin 2A in Terms of A****cos 2A in Terms of A****tan 2A in Terms of A****sin 2A in Terms of tan A****cos 2A in Terms of tan A****Trigonometric Functions of A in Terms of cos 2A****sin 3A in Terms of A****cos 3A in Terms of A****tan 3A in Terms of A****Multiple Angle Formulae**

**11 and 12 Grade Math**

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