Sin 2A Formula – Proof and Examples

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 = \(24\over 25\)

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