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
11 and 12 Grade Math
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