# cos 2A in Terms of A

We will learn to express trigonometric function of cos 2A in terms of A. We know if A is a given angle then 2A is known as multiple angles.

How to proof the formula of cos 2A is equals cos$$^{2}$$ A - sin$$^{2}$$ A?

Or

How to proof the formula of cos 2A is equals 1 - 2 sin$$^{2}$$ A?

Or

How to proof the formula of cos 2A is equals 2 cos$$^{2}$$ A - 1?

We know that for two real numbers or angles A and B,

cos (A + B) = cos A cos B - sin A sin B

Now, putting B = A on both sides of the above formula we get,

cos (A + A) = cos A cos A - sin A sin A

cos 2A = cos$$^{2}$$ A - sin$$^{2}$$ A

⇒ cos 2A = cos$$^{2}$$ A - (1 - cos$$^{2}$$ A), [since we know that sin$$^{2}$$ θ = 1 - cos$$^{2}$$ θ]

⇒ cos 2A = cos$$^{2}$$ A - 1 + cos$$^{2}$$ A,

cos 2A = 2 cos$$^{2}$$ A - 1

⇒ cos 2A = 2 (1 - sin$$^{2}$$ A) - 1, [since we know that cos$$^{2}$$ θ = 1 - sin$$^{2}$$ θ]

⇒ cos 2A = 2 - 2 sin$$^{2}$$ A - 1

cos 2A = 1 - 2 sin$$^{2}$$ A

Note:

(i) From cos 2A = 2 cos$$^{2}$$ A - 1 we get, 2 cos$$^{2}$$ A = 1 + cos 2A

and from cos 2A = 1 - 2 sin$$^{2}$$ A we get,  2 sin$$^{2}$$A = 1 - cos 2A

(ii) In the above formula we should note that the angle on the R.H.S. is half of the angle on L.H.S. Therefore, cos 120° = cos$$^{2}$$ 60° - sin$$^{2}$$ 60°.

(iii) The above formulae is also known as double angle formulae for cos 2A.

Now, we will apply the formula of multiple angle of cos 2A in terms of A to solve the below problems.

1. Express cos 4A in terms of sin 2A and cos 2A

Solution:

cos 4A

= cos (2 ∙ 2A)

= cos$$^{2}$$ (2A) - sin$$^{2}$$ (2A)



2. Express cos 4β in terms of sin 2β

Solution:

cos 4β

= cos (2 ∙ 2β)

= 1 - 2 sin$$^{2}$$ (2β)

3. Express cos 4θ in terms of cos 2θ

Solution:

cos 4θ

= cos 2 ∙ 2θ

= 2 cos$$^{2}$$ (2θ) – 1

4. Express cos 4A in term of cos A.

Solution:

cos 4A = cos (2 ∙ 2A) = 2 cos$$^{2}$$ (2A) - 1

⇒ cos 4A = 2(2 cos 2A - 1)$$^{2}$$ - 1

⇒ cos 4A = 2(4 cos$$^{4}$$ A - 4 cos$$^{2}$$ A + 1) - 1

⇒ cos 4A =  8 cos$$^{4}$$ A – 8 cos$$^{2}$$ A + 1

More solved examples on cos 2A in terms of A.

5. If sin A = $$\frac{3}{5}$$ find the values of cos 2A.

Solution:

Given, sin A = $$\frac{3}{5}$$

cos 2A

= 1 - 2 sin$$^{2}$$ A

= 1 - 2 ($$\frac{3}{5}$$)$$^{2}$$

= 1 - 2 ($$\frac{9}{25}$$)

= 1 - $$\frac{18}{25}$$

= $$\frac{25 - 18}{25}$$

= $$\frac{7}{25}$$

6. Prove that cos 4x = 1 - sin$$^{2}$$ x cos$$^{2}$$ x

Solution:

L.H.S. = cos 4x

= cos (2 × 2x)

= 1 - 2 sin$$^{2}$$ 2x, [Since, cos 2A = 1 - 2 sin$$^{2}$$ A]

= 1 - 2 (2 sin x cos x)$$^{2}$$

= 1 - 2 (4 sin$$^{2}$$ x cos$$^{2}$$ x)

= 1 - 8 sin$$^{2}$$ x cos$$^{2}$$ x = R.H.S.           Proved