Lesson 9.3 : Double-Angle, Half-Angle, and Reduction Formulas

New Concept

This lesson introduces powerful trigonometric identities: double-angle, half-angle, and reduction formulas. You'll use these to find exact values, simplify expressions by reducing powers, and verify more complex identities, greatly expanding your problem-solving toolkit.

What’s next

First, you'll master the double-angle formulas. Get ready for a sequence of interactive examples and practice cards to find exact values and verify identities.

Property

The double-angle formulas are a special case of the sum formulas, where $\alpha = \beta$. They are summarized as follows:

To find exact values, first draw a triangle from the given information. Then, determine the correct double-angle formula, substitute values from the triangle, and simplify.

Examples

• Given that $\cos \theta = \frac{5}{13}$ and $\theta$ is in quadrant I, we find $\sin \theta = \frac{12}{13}$. Then, $\sin(2\theta) = 2 \sin \theta \cos \theta = 2(\frac{12}{13})(\frac{5}{13}) = \frac{120}{169}$.
• Given that $\sin \theta = \frac{4}{5}$ and $\theta$ is in quadrant II, we find $\cos \theta = -\frac{3}{5}$. We can use the formula $\cos(2\theta) = 1 - 2\sin^2 \theta = 1 - 2(\frac{4}{5})^2 = 1 - 2(\frac{16}{25}) = 1 - \frac{32}{25} = -\frac{7}{25}$.
• Given that $\tan \theta = \frac{1}{2}$ and $\theta$ is in quadrant III, we can find $\tan(2\theta)$ directly. $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2(\frac{1}{2})}{1 - (\frac{1}{2})^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

Explanation

These formulas are shortcuts derived from sum formulas, used when an angle is added to itself. They let you find trigonometric values for a doubled angle, like $2\theta$, using the known values of the original angle $\theta$.

Property

The reduction formulas, also called power-reducing formulas, are derived from the double-angle formulas. They allow us to rewrite even powers of sine or cosine in terms of the first power of cosine. The formulas are summarized as follows:

Examples

• To write an equivalent expression for $\sin^4 x$ without powers greater than 1, we apply the reduction formula twice: $\sin^4 x = (\sin^2 x)^2 = (\frac{1 - \cos(2x)}{2})^2 = \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x)) = \frac{1}{4} - \frac{1}{2}\cos(2x) + \frac{1}{4}(\frac{1 + \cos(4x)}{2}) = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)$.
• Use the power-reducing formulas to prove $\sin^2 x \cos^2 x = \frac{1 - \cos(4x)}{8}$. We start with the left side: $(\frac{1 - \cos(2x)}{2})(\frac{1 + \cos(2x)}{2}) = \frac{1 - \cos^2(2x)}{4} = \frac{\sin^2(2x)}{4} = \frac{1}{4}(\frac{1 - \cos(4x)}{2}) = \frac{1 - \cos(4x)}{8}$.
• Rewrite $\cos^2(3x)$ with no exponent higher than 1. Using the formula for $\cos^2 \theta$ with $\theta = 3x$, we get $\cos^2(3x) = \frac{1 + \cos(2(3x))}{2} = \frac{1 + \cos(6x)}{2}$.

Explanation

These formulas reduce the power of a trigonometric expression. They transform terms with exponents, like $\cos^2 \theta$, into equivalent expressions without powers greater than one, which is essential for simplifying problems in higher-level math like calculus.

Property

The half-angle formulas are derived from the reduction formulas and are used to find exact values for an angle that is half the size of a special angle. The $\pm$ sign indicates that you must choose the correct sign based on the quadrant in which the half-angle $\frac{\alpha}{2}$ terminates.

Examples

• To find the exact value of $\cos(22.5^\circ)$, use the half-angle formula for cosine with $\alpha = 45^\circ$. Since $22.5^\circ$ is in quadrant I, the result is positive: $\cos(22.5^\circ) = \sqrt{\frac{1 + \cos(45^\circ)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$.
• Given $\tan \alpha = \frac{24}{7}$ and $\alpha$ is in quadrant III, find $\sin(\frac{\alpha}{2})$. First, find $\cos \alpha = -\frac{7}{25}$. Since $180^\circ < \alpha < 270^\circ$, then $90^\circ < \frac{\alpha}{2} < 135^\circ$, which is in quadrant II, so sine is positive. $\sin(\frac{\alpha}{2}) = \sqrt{\frac{1 - (-\frac{7}{25})}{2}} = \sqrt{\frac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
• Given $\sin \alpha = -\frac{3}{5}$ and $\alpha$ is in quadrant IV, find $\cos(\frac{\alpha}{2})$. First, $\cos \alpha = \frac{4}{5}$. Since $270^\circ < \alpha < 360^\circ$, then $135^\circ < \frac{\alpha}{2} < 180^\circ$, which is in quadrant II, so cosine is negative. $\cos(\frac{\alpha}{2}) = -\sqrt{\frac{1 + \frac{4}{5}}{2}} = -\sqrt{\frac{\frac{9}{5}}{2}} = -\sqrt{\frac{9}{10}} = -\frac{3}{\sqrt{10}} = -\frac{3\sqrt{10}}{10}$.

Explanation

These formulas find trig values for half an angle, like $22.5^\circ$, by using the trig values of the full angle, $45^\circ$. The most important step is choosing the positive or negative sign based on the quadrant of the half angle.