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# Double Angle Formula

$\sin (A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B$

By putting $B = A$, we have

$\sin (A + A) = \sin A \cdot \cos A + \cos A \cdot \sin A$

$\Longrightarrow \quad \sin 2A = 2 \cdot \sin A \cdot \cos A$

Thus,

 $\sin 2A = 2 \cdot \sin A \cdot \cos A$

In Trigonometry , From Addition Formula, we have

$\cos (A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B$.

By putting $B = A$, we have

$\cos (A + A) = \cos A \cdot \cos A - \sin A \cdot \sin A$

$\Longrightarrow \quad \cos 2A = \cos^2 A - \sin^2 A$. ------------- (1)

We know that

$\cos^2 A = 1 - \sin^2 A$ and $\sin^2 A = 1 - \cos^2 A$.

Therefore, from (1), we have

$\cos 2A = \cos^2 A - \sin^2 A$

$= \cos^2 A - (1 - \cos^2 A)$

$= \cos^2 A - 1 + \cos^2 A$

$= 2 \cdot \cos^2 A - 1$.

Again $\cos^2 A = 1 - \sin^2 A$ implies,

$\cos 2A = 2 \cdot (1 - \sin^2 A) - 1$

$= 2 - 2 \cdot \sin^2 A - 1$

$= 1 - 2 \cdot \sin^2 A$.

Thus, we have

 $\cos 2A = \cos^2 A - \sin^2 A$$= 2 \cdot \cos^2 A - 1$$= 1 - 2 \cdot \sin^2 A$

## Double Angle Identities

We know that

$\cos^2 A + \sin^2 A$ $= 1$

Now

$\sin 2A$ $=$ $\frac{\sin 2A}{1}$

$=$ $\frac{2 \cdot \sin A \cdot \cos A}{\cos^2 A + \sin^2 A}$

$=$ $\frac{(\frac{2 \cdot \sin A \cdot \cos A}{\cos^2 A})}{(\frac{\cos^2 A + \sin^2 A}{\cos^2 A})}$

$=$ $\frac{(\frac{2 \cdot \sin A}{\cos A})}{(\frac{\cos^2 A}{\cos^2 A} + \frac{\sin^2 A}{\cos^2 A})}$

$=$ $\frac{2 \cdot \tan A}{1 + \tan^2 A}$.

Thus,

 $\sin 2A =$ $\frac{2 \cdot \tan A}{1 + \tan^2 A}$

## Given that $An Alpha =$ $frac{5}{12}$. Find out $sin 2a$.

Solution

We know that

$\sin 2\alpha =$ $\frac{2 \cdot \tan \alpha}{1 + \tan^2 \alpha}$

$=$ $\frac{2 \cdot (\frac{5}{12})}{1 + (\frac{5}{12})^2}$

$=$ $\frac{(\frac{5}{6})}{1 + \frac{25}{144}}$

$=$ $\frac{(\frac{5}{6})}{(\frac{169}{144})}$

$=$ $\frac{5}{6} \times \frac{144}{169}$

$=$ $\frac{5 \times 24}{169}$

$=$ $\frac{120}{169}$.

## Expressing Trigonometric Ratios for Double-angle Formulas

In Trigonometric Ratios, We know that

$\sin^2 A + \cos^2 A$ $= 1$

We have

$2 \cdot \sin A \cdot \cos A = \sin 2A$.

Then

$\sin^2 A + \cos^2 A + 2 \cdot \sin A \cdot \cos A = 1 + \sin 2A$

$\Longrightarrow \quad (\sin A + \cos A)^2 = 1 + \sin 2A$

$\Longrightarrow \quad \sin A + \cos A = \pm \sqrt{1 + \sin 2A}$. ---------------- (1)

Similarly,

$\sin^2 A + \cos^2 A - 2 \cdot \sin A \cdot \cos A = 1 - \sin 2A$

$\Longrightarrow \quad (\sin A - \cos A)^2 = 1 - \sin 2A$

$\Longrightarrow \quad \sin A - \cos A = \pm \sqrt{1 - \sin 2A}$. ---------------- (2)

First by adding (1) and (2), then by subtracting them, we obtain

$2 \sin A = \pm \sqrt{1 + \sin 2A} \pm \sqrt{1 - \sin 2A}$, and

$2 \cos A = \pm \sqrt{1 + \sin 2A} \mp \sqrt{1 - \sin 2A}$

Thus, we have

$\sin A =$ $\frac{1}{2}$ $(\pm \sqrt{1 + \sin 2A} \pm \sqrt{1 - \sin 2A})$ ---------- (3)

$\cos A =$ $\frac{1}{2}$ $(\pm \sqrt{1 + \sin 2A} \mp \sqrt{1 - \sin 2A})$. ---------- (4)

The other ratios of $A$ can then easily be obtained.

In each of the formulae (3) and (4) there are two ambiguous signs. The appropriate sign to be taken is dependent on the measurement of the angle $A$.