Double Angle, Half Angle, and Reduction Formulas: Learn It 4

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Double Angle, Half Angle, and Reduction Formulas: Learn It 4

Using Half-Angle Formulas to Find Exact Values

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\theta[/latex] with [latex]\frac{\alpha }{2}[/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\sin \left(\frac{\alpha }{2}\right)[/latex]. Note that the half-angle formulas are preceded by a [latex]\pm[/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\frac{\alpha }{2}[/latex] terminates.

The half-angle formula for sine is derived as follows:

[latex]\begin{align}{\sin }^{2}\theta &=\frac{1-\cos \left(2\theta \right)}{2}\\ {\sin }^{2}\left(\frac{\alpha }{2}\right)&=\frac{1-\left(\cos 2\cdot \frac{\alpha }{2}\right)}{2} \\ &=\frac{1-\cos \alpha }{2} \\ \sin \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \end{align}[/latex]

To derive the half-angle formula for cosine, we have

[latex]\begin{align}{\cos }^{2}\theta &=\frac{1+\cos \left(2\theta \right)}{2}\\ {\cos }^{2}\left(\frac{\alpha }{2}\right)&=\frac{1+\cos \left(2\cdot \frac{\alpha }{2}\right)}{2} \\ &=\frac{1+\cos \alpha }{2} \\ \cos \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \end{align}[/latex]

For the tangent identity, we have

[latex]\begin{align}{\tan }^{2}\theta &=\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \\ {\tan }^{2}\left(\frac{\alpha }{2}\right)&=\frac{1-\cos \left(2\cdot \frac{\alpha }{2}\right)}{1+\cos \left(2\cdot \frac{\alpha }{2}\right)} \\ &=\frac{1-\cos \alpha }{1+\cos \alpha }\hfill \\ \tan \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \end{align}[/latex]

half angle formulas

[latex]\begin{align}\sin \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \\ \text{ } \\ \cos \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \\ \text{ } \\ \tan \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \\ &=\frac{\sin \alpha }{1+\cos \alpha } \\ &=\frac{1-\cos \alpha }{\sin \alpha }\end{align}[/latex]

Find [latex]\sin \left({15}^{\circ }\right)[/latex] using a half-angle formula.

How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

Draw a triangle to represent the given information.
Determine the correct half-angle formula.
Substitute values into the formula based on the triangle.
Simplify.

Given that [latex]\tan \alpha =\frac{8}{15}[/latex] and [latex]\alpha[/latex] lies in quadrant III, find the exact value of the following:

[latex]\sin \left(\frac{\alpha }{2}\right)[/latex]
[latex]\cos \left(\frac{\alpha }{2}\right)[/latex]
[latex]\tan \left(\frac{\alpha }{2}\right)[/latex]

Using the given information, we can draw the triangle. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\sin \alpha =-\frac{8}{17}[/latex] and [latex]\cos \alpha =-\frac{15}{17}[/latex].

Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.

Before we start, we must remember that, if [latex]\alpha[/latex] is in quadrant III, then [latex]180^\circ <\alpha <270^\circ[/latex], so [latex]\frac{180^\circ }{2}<\frac{\alpha }{2}<\frac{270^\circ }{2}[/latex]. This means that the terminal side of [latex]\frac{\alpha }{2}[/latex] is in quadrant II, since [latex]90^\circ <\frac{\alpha }{2}<135^\circ[/latex].To find [latex]\sin \frac{\alpha }{2}[/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle and simplify.
[latex]\begin{align} \sin \frac{\alpha }{2}&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \\ &=\pm \sqrt{\frac{1-\left(-\frac{15}{17}\right)}{2}} \\ &=\pm \sqrt{\frac{\frac{32}{17}}{2}} \\ &=\pm \sqrt{\frac{32}{17}\cdot \frac{1}{2}} \\ &=\pm \sqrt{\frac{16}{17}} \\ &=\pm \frac{4}{\sqrt{17}} \\ &=\frac{4\sqrt{17}}{17} \end{align}[/latex]

We choose the positive value of [latex]\sin \frac{\alpha }{2}[/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.
To find [latex]\cos \frac{\alpha }{2}[/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle, and simplify.
[latex]\begin{align} \cos \frac{\alpha }{2}&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \\ &=\pm \sqrt{\frac{1+\left(-\frac{15}{17}\right)}{2}} \\ &=\pm \sqrt{\frac{\frac{2}{17}}{2}} \\ &=\pm \sqrt{\frac{2}{17}\cdot \frac{1}{2}} \\ &=\pm \sqrt{\frac{1}{17}} \\ &=-\frac{\sqrt{17}}{17}\end{align}[/latex]

We choose the negative value of [latex]\cos \frac{\alpha }{2}[/latex] because the angle is in quadrant II because cosine is negative in quadrant II.
To find [latex]\tan \frac{\alpha }{2}[/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle and simplify.
[latex]\begin{align} \tan \frac{\alpha }{2}&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \\ &=\pm \sqrt{\frac{1-\left(-\frac{15}{17}\right)}{1+\left(-\frac{15}{17}\right)}} \\ &=\pm \sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}} \\ &=\pm \sqrt{\frac{32}{2}} \\ &=-\sqrt{16} \\ &=-4 \end{align}[/latex]

We choose the negative value of [latex]\tan \frac{\alpha }{2}[/latex] because [latex]\frac{\alpha }{2}[/latex] lies in quadrant II, and tangent is negative in quadrant II.

Given that [latex]\sin \alpha =-\frac{4}{5}[/latex] and [latex]\alpha[/latex] lies in quadrant IV, find the exact value of [latex]\cos \left(\frac{\alpha }{2}\right)[/latex].