Here you will learn what is the formula of sin 2A in terms of sin and cos and also in terms of tan with proof and examples.
Let’s begin –
Sin 2A Formula
(i) In Terms of Cos and Sin :
Sin 2A = 2 sin A cos A
Proof :
We have,
Sin (A + B) = sin A cos B + cos A sin B
Replacing B by A,
\(\implies\) sin 2A = sin A cos A + cos A sin A
\(\implies\) sin 2A = 2 sin A cos A
We can also write above relation in terms of angle A/2, just replace A by A/2, we get
sin A = \(2 sin ({A\over 2}) cos ({A\over 2})\)
(ii) Sin 2A Formula in Terms of Tan :
Sin 2A = \(2 tan A\over 1 + tan^2 A\)
Proof :
We have,
sin 2A = 2 sin A cos A
\(\implies\) sin 2A = \(2 sin A cos A\over sin^2 A + cos^2 A\)
[ \(\because\) \(sin^2 A + cos^2 A\) = 1 ]
Now, Dividing numerator and denominator by \(cos^2 A\),
\(\implies\) sin 2A = \({2sin A cos A\over cos^2 A}\over {sin^2 A + cos^2 A\over cos^2 A}\)
\(\implies\) sin 2A = \(2 tan A\over 1 + tan^2 A\)
We can also write above relation in terms of angle A/2, just replace A by A/2, we get
sin A = \(2 tan ({A\over 2})\over 1 + tan^2 ({A\over 2})\)
Example : Find the value of Sin 120 ?
Solution : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\)
By using above formula,
sin 120 = 2 sin 60 cos 60 = 2 \(\times\) \(\sqrt{3}\over 2\) \(\times\) \(1\over 2\)
\(\implies\) sin 120 = \(\sqrt{3}\over 2\)
Example : If sin A = \(3\over 5\), where 0 < A < 90, find the value of sin 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,
sin 2A = 2 sin A cos A = 2 \(\times\) \(3\over 5\) \(\times\) \(4\over 5\)
\(\implies\) sin 2A = \(24\over 25\)
