\( \DeclareMathOperator{cosec}{cosec} \)
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Double angle identities are a special case of the compound angle identities. They can be derived by setting both angles in the compound angle identities to be the same. These identities are particularly useful in simplifying expressions involving trigonometric functions and solving trigonometric equations.
The double angle identities for sine and cosine are derived from the sum formulas for sine and cosine, by setting \( A = B = \theta \):
$$ \sin(2\theta) = \sin(\theta + \theta) = \sin \theta \cos \theta + \cos \theta \sin \theta = 2 \sin \theta \cos \theta $$
$$ \cos(2\theta) = \cos(\theta + \theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos^2 \theta - \sin^2 \theta $$
The double angle identity for tangent is derived in a similar way from the sum formula for tangent:
$$ \tan(2\theta) = \frac{\tan \theta + \tan \theta}{1 - \tan \theta \tan \theta} = \frac{2 \tan \theta}{1 - \tan^2 \theta} $$
Let's look at an example using the double angle identity for sine:
Example: Simplify \( \sin(2x) \) when \( \sin(x) = \frac{1}{2} \)
Using the double angle identity for sine, we have:
$$ \sin(2x) = 2 \sin(x) \cos(x) $$
Since \( \sin^2(x) + \cos^2(x) = 1 \), we can solve for \( \cos(x) \):
$$ \cos(x) = \sqrt{1 - \sin^2(x)} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$
Now we substitute \( \sin(x) \) and \( \cos(x) \) into the double angle identity:
$$ \sin(2x) = 2 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} $$
Therefore, \( \sin(2x) \) simplifies to \( \frac{\sqrt{3}}{2} \) when \( \sin(x) = \frac{1}{2} \).
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