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*

**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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