Here you will learn what is the formula of cos 2A in terms of sin and cos and also in terms of tan with proof and examples.
Let’s begin –
Cos 2A Formula :
(i) In Terms of Cos and Sin
Given below are all the formulas for cos 2A.
(i) cos 2A = \(cos^2 A\) – \(sin^2 A\)
(ii) cos 2A = \(2cos^2 A – 1\) or, 1 + cos 2A = \(2cos^2 A\)
(iii) cos 2A = \(1 – 2sin^2 A\) or, 1 – cos 2A = \(2sin^2 A\)
Proof :
(i) We have,
Cos (A + B) = cos A cos B – sin A sin B
Replacing B by A,
\(\implies\) cos 2A = cos A cos A + sin A sin A
\(\implies\) cos 2A = \(cos^2 A\) – \(sin^2 A\)
(ii) We have,
cos 2A = \(cos^2 A\) – \(sin^2 A\)
\(\implies\) cos 2A = \(cos^2 A\) – \(1 – cos^2 A\)
\(\implies\) cos 2A = \(2cos^2 A – 1\)
Again, cos 2A = \(2cos^2 A – 1\)
\(\implies\) 1 + cos 2A = \(2cos^2 A\)
(iii) We have,
cos 2A = \(cos^2 A\) – \(sin^2 A\)
\(\implies\) cos 2A = \(1 – sin^2 A\) – \(sin^2 A\)
\(\implies\) cos 2A = \(1- 2sin^2 A\)
Again, cos 2A = \(1- 2sin^2 A\)
\(\implies\) 1 – cos 2A = \(2ain^2 A\)
We can also write above relation in terms of angle A/2, just replace A by A/2, we get
(i) cos A = \(cos^2 ({A\over 2})\) – \(sin^2 ({A\over 2})\)
(ii) cos A = \(2cos^2 ({A\over 2}) – 1\) or, 1 + cos A = \(2cos^2 ({A\over 2})\)
(iii) cos A = \(1 – 2sin^2 ({A\over 2})\) or, 1 – cos A = \(2sin^2 ({A\over 2})\)
(ii) Cos 2A Formula in Terms of Tan :
Cos 2A = \(1 – tan^2 A\over 1 + tan^2 A\)
Proof :
We have,
cos 2A = \(cos^2 A\) – \(sin^2 A\)
\(\implies\) cos 2A = \(cos^2 A – sin^2 A\over sin^2 A + cos^2 A\)
[ \(\because\) \(sin^2 A + cos^2 A\) = 1 ]
Now, Dividing numerator and denominator by \(cos^2 A\),
\(\implies\) cos 2A = \({cos^2 A – sin^2 A\over cos^2 A}\over {sin^2 A + cos^2 A\over cos^2 A}\)
\(\implies\) cos 2A = \(1 – tan^2 A\over 1 + tan^2 A\)
We can also write above relation in terms of angle A/2, just replace A by A/2, we get
cos A = \(1 – tan^2 ({A\over 2})\over 1 + tan^2 ({A\over 2})\)
Example : Find the value of Cos 120 ?
Solution : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\)
By using above formula,
cos 120 = \(cos^2 60\) – \(sin^2 60\) = \(1\over 4\) – \(3\over 4\)
\(\implies\) cos 120 = \(-1\over 2\)
Example : If sin A = \(3\over 5\), where 0 < A < 90, find the value of cos 2A ?
Solution : We have,
sin A = \(3\over 5\) where 0 < A < 90 degrees
\(\therefore\) \(cos^2 A\) = 1 – \(sin^2 A\)
\(\implies\) cos A = \(\sqrt{1 – sin^2 A}\) = \(\sqrt{1 – {9\over 25}}\) = \(4\over 5\)
By using above formula,
cos 2A = \(cos^2 A\) – \(sin^2 A\) = \(16\over 25\) – \(9\over 25\)
\(\implies\) cos 2A = \(7\over 25\)
