9.4: Half-Angle and Power Reduction Identities

Example \(\PageIndex{1}\)

Write an equivalent expression for \(\cos^4\left(x\right)\) that does not involve any powers of sine or cosine greater than \(1\).

Solution

We will apply the Reduction Identity for cosine twice.\[\begin{array}{rclcl}\cos^4\left(x\right) & = & \left(\cos^2\left(x\right)\right)^2 & & \\[6pt] & = & \left(\dfrac{1+\cos(2x)}{2}\right)^2 & \quad & \left(\text{Reduction Identity}\right) \\[6pt] & = & \dfrac{1}{4}\left(1+2\cos(2x)+\cos^2(2x)\right) & \quad & \left(\text{distribution}\right) \\[6pt] & = & \dfrac{1}{4}+\dfrac{1}{2}\cos(2x)+\dfrac{1}{4}\left(\dfrac{1+\cos(4x)}{2}\right) & \quad & \left(\text{distribution and Reduction Identity}\right) \\[6pt] & = & \dfrac{1}{4}+\dfrac{1}{2}\cos(2x)+\dfrac{1}{8}+\dfrac{1}{8}\cos(4x) & \quad & \left(\text{distribution}\right) \\[6pt] & = & \dfrac{3}{8}+\dfrac{1}{2}\cos(2x)+\dfrac{1}{8}\cos(4x) & \quad & \left(\text{combining like terms}\right) \\[6pt] \end{array}\nonumber\]

Example \(\PageIndex{2}\)

Prove\[\sin^3(2x)=\dfrac{1}{2}\sin(2x)\left[1-\cos(4x)\right].\nonumber\]

Solution

We will work on simplifying the left side of the equation.\[\begin{array}{rclcl}\text{LHS} & = & \sin^3(2x) & & \\[6pt] & = & \sin(2x)\cdot\sin^2(2x) & & \\[6pt] & = & \sin(2x)\left[\dfrac{1-\cos(4x)}{2}\right] & \quad & \left(\text{Power Reduction Identity}\right) \\[6pt] & = & \sin(2x)\left(\dfrac{1}{2}\right)[1-\cos(4x)] & & \\[6pt] & = & \dfrac{1}{2}\sin(2x)[1-\cos(4x)] & & \\[6pt] & = & \text{RHS} & & \\[6pt] \end{array}\nonumber\]

Example \(\PageIndex{3}\)

Find the exact value of \(\sin\left(15^{\circ}\right)\).

Solution

Since \(15^{\circ}=\dfrac{30^{\circ}}{2}\), we use the Half-Angle Identity for sine:\[\begin{array}{rclcl}\sin\left(15^{\circ}\right) & = & \sin\left(\dfrac{30^{\circ}}{2}\right) & & \\[6pt] & = & +\sqrt{\dfrac{1-\cos\left(30^{\circ}\right)}{2}} & \quad & \left(\text{since }15^{\circ}\in\mathrm{QI}\text{, }\sin\left(15^{\circ}\right)\text{ is positive}\right) \\[6pt] & = & \sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}} & & \\[6pt] & = & \sqrt{\dfrac{2-\sqrt{3}}{4}} & & \\[6pt] & = & \dfrac{\sqrt{2-\sqrt{3}}}{2} & & \\[6pt] \end{array}\nonumber\]Remember that we can check the answer with a graphing calculator.

Example \(\PageIndex{4}\)

Given that \(\tan\left(\alpha\right)=\dfrac{8}{15}\) and \(\alpha\) lies in quadrant III, find the exact value of the following:

1. \(\sin\left(\dfrac{\alpha}{2}\right)\)
2. \(\cos\left(\dfrac{\alpha}{2}\right)\)
3. \(\tan\left(\dfrac{\alpha}{2}\right)\)

Solutions

Using the given information, we draw the triangle shown below.

Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate\[\sin\left(\alpha\right)=-\dfrac{8}{17}\nonumber\]and\[\cos\left(\alpha\right)=-\dfrac{15}{17}.\nonumber\]Before we start, we must remember that if \(\alpha\) is in quadrant III, then \(\pi\lt\alpha\lt\dfrac{3\pi}{2}\), so \(\dfrac{\pi}{2}\lt\dfrac{\alpha}{2}\lt\dfrac{3\pi}{4}\). This means that the terminal side of \(\dfrac{\alpha}{2}\) is in quadrant II.

1. To find \(\sin\left(\dfrac{\alpha}{2}\right)\), we begin by writing the Half-Angle Identity for sine. Then, we substitute the value of the cosine we found from the triangle above and simplify.\[\begin{array}{rclcl}\sin\left(\dfrac{\alpha}{2}\right) & = & \pm\sqrt{\dfrac{1-\cos\left(\alpha\right)}{2}} & \quad & \left(\text{Half-Angle Identity}\right) \\[6pt] & = & \pm\sqrt{\dfrac{1-\left(-\dfrac{15}{17}\right)}{2}} & & \\[6pt] & = & \pm\sqrt{\dfrac{\dfrac{32}{17}}{2}} & & \\[6pt] & = & \pm\sqrt{\dfrac{32}{34}} & & \\[6pt] & = & \pm\sqrt{\dfrac{16}{17}} & & \\[6pt] & = & \pm\dfrac{4}{\sqrt{17}} & & \\[6pt] & = & \dfrac{4}{\sqrt{17}} & \quad & \left(\dfrac{\alpha}{2}\in\mathrm{QII}\right) \\[6pt] \end{array}\nonumber\]We choose the positive value of \(\sin\left(\dfrac{\alpha}{2}\right)\) because the angle terminates in quadrant II and sine is positive in quadrant II.
2. To find \(\cos\left(\dfrac{\alpha}{2}\right)\), we will write the Half-Angle Identity for cosine, substitute the value of the cosine we found from the triangle, and simplify.\[\begin{array}{rclcl}\cos\left(\dfrac{\alpha}{2}\right) & = & \pm\sqrt{\dfrac{1+\cos\left(\alpha\right)}{2}} & \quad & \left(\text{Half-Angle Identity}\right) \\[6pt] & = & \pm\sqrt{\dfrac{1+\left(-\dfrac{15}{17}\right)}{2}} & & \\[6pt] & = & \pm\sqrt{\dfrac{\dfrac{2}{17}}{2}} & & \\[6pt] & = & \pm\sqrt{\dfrac{2}{34}} & & \\[6pt] & = & \pm\sqrt{\dfrac{1}{17}} & & \\[6pt] & = & -\dfrac{1}{\sqrt{17}} & \quad & \left(\dfrac{\alpha}{2}\in\mathrm{QII}\right) \\[6pt] \end{array}\nonumber\]We choose the negative value of \(\cos\left(\dfrac{\alpha}{2}\right)\) because the angle is in quadrant II and the cosine is negative in quadrant II.
3. To find \(\tan\left(\dfrac{\alpha}{2}\right)\), we write the Half-Angle Identity for the tangent. Again, we substitute the value of the cosine we found from the triangle and simplify.\[\begin{array}{rclcl}\tan\left(\dfrac{\alpha}{2}\right) & = & \pm\sqrt{\dfrac{1-\cos\left(\alpha\right)}{1+\cos\left(\alpha\right)}} & \quad & \left(\text{Half-Angle Identity}\right) \\[6pt] & = & \pm\sqrt{\dfrac{1-\left(-\dfrac{15}{17}\right)}{1+\left(-\dfrac{15}{17}\right)}} & & \\[6pt] & = & \pm\sqrt{\dfrac{\dfrac{32}{17}}{\dfrac{2}{17}}} & & \\[6pt] & = & \pm\sqrt{\dfrac{32}{2}} & & \\[6pt] & = & -\sqrt{16} & \quad & \left(\dfrac{\alpha}{2}\in\mathrm{QII}\right) \\[6pt] & = & -4 & & \\[6pt] \end{array}\nonumber\]We choose the negative value of \(\tan\left(\dfrac{\alpha}{2}\right)\) because \(\dfrac{\alpha}{2}\) lies in quadrant II, and tangent is negative in quadrant II.

Example \(\PageIndex{5}\)

A bicycle ramp is constructed for high-level competition with an angle of \(\theta\) formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If \(\tan\left(\theta\right)=\dfrac{5}{3}\) for higher-level competition, what is the measurement of the angle for novice competition?

Solution

Since the angle for novice competition measures half the steepness of the angle for the high-level competition, and \(\tan\left(\theta\right)=\dfrac{5}{3}\) for high competition, we can find \(\cos\left(\theta\right)\) from the right triangle and the Pythagorean Theorem so that we can use the Half-Angle Identities (see the figure below).

We see that \(\cos\left(\theta\right)=\dfrac{3}{\sqrt{34}}\). We can use the Half-Angle Identity for tangent:\[\tan\left(\dfrac{\theta}{2}\right)=\sqrt{\dfrac{1-\cos\left(\theta\right)}{1+\cos\left(\theta\right)}}.\nonumber\]Since \(\tan\left(\theta\right)\) is in the first quadrant, so is \(\tan\left(\dfrac{\theta}{2}\right)\).\[\begin{array}{rcl}\tan\left(\dfrac{\theta}{2}\right) & = & \sqrt{\dfrac{1-\dfrac{3}{\sqrt{34}}}{1+\dfrac{3}{\sqrt{34}}}} \\[6pt] & = & \sqrt{\dfrac{\sqrt{34}-3}{\sqrt{34}+3}} \\[6pt] & \approx & 0.57 \\[6pt] \end{array}\nonumber\]We can take the inverse tangent to find the angle: \(\tan^{-1}(0.57)\approx 29.7^{\circ}\). So, the angle of the ramp for novice competition is \(\approx 29.7^{\circ}\).

Example \(\PageIndex{6}\)

Graph the function.\[y=3\sin^2\left(\frac{x}{2}\right)\nonumber\]

Solution

We begin by recognizing this as the square of a Half-Angle Identity.\[\begin{array}{rcl}y & = & 3\sin^2\left(\dfrac{x}{2}\right)\\[6pt] & = & 3\left[\pm\sqrt{\dfrac{1-\cos\left(x\right)}{2}}\right]^2 \\[6pt] & = & 3\left(\dfrac{1-\cos\left(x\right)}{2}\right) \\[6pt] & = & -\dfrac{3}{2}\cos\left(x\right)+\dfrac{3}{2} \\[6pt] \end{array}\nonumber\]We see this is a reflected cosine function with amplitude \(\dfrac{3}{2}\), period \(2\pi\), step size \(\dfrac{\pi}{2}\), phase shift 0, and vertical shift \(\dfrac{3}{2}\). Using all this information, we obtain the following graph.