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3.6: Half-Angle Identities

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Apply the half angle identities to expressions, equations and other identities.
  • Use the half angle identities to find the exact value of trigonometric functions for certain angles.

Just as there are double angle identities, there are also half angle identities. For example: \begin{align*}\sin \frac{1}{2} a\end{align*} can be found in terms of the angle \begin{align*}``a''\end{align*}. Recall that \begin{align*}\frac{1}{2} a\end{align*} and \begin{align*}\frac{a}{2}\end{align*} are the same thing and will be used interchangeably throughout this section.

Deriving the Half Angle Formulas

In the previous lesson, one of the formulas that was derived for the cosine of a double angle is: \begin{align*}\cos 2 \theta = 1 - 2 \sin^2 \theta\end{align*}. Set \begin{align*}\theta = \frac{\alpha}{2}\end{align*}, so the equation above becomes \begin{align*}\cos 2 \frac{\alpha}{2} = 1 - 2 \sin^2 \frac{\alpha}{2}\end{align*}.

Solving this for \begin{align*}\sin \frac{\alpha}{2}\end{align*}, we get:

\begin{align*}\cos 2 \frac{\alpha}{2} & = 1 - 2 \sin^2 \frac{\alpha}{2} \\ \cos \alpha & = 1 - 2 \sin^2 \frac{\alpha}{2} \\ 2 \sin^2 \frac{\alpha}{2} & = 1 - \cos \alpha \\ \sin^2 \frac{\alpha}{2} & = \frac{1 - \cos \alpha}{2} \\ \sin \frac{\alpha}{2} & = \pm \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*}

\begin{align*}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*} if \begin{align*}\frac{\alpha}{2}\end{align*} is located in either the first or second quadrant.

\begin{align*}\sin \frac{\alpha}{2} = - \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*} if \begin{align*}\frac{\alpha}{2}\end{align*} is located in the third or fourth quadrant.

Example 1: Determine the exact value of \begin{align*}\sin 15^\circ\end{align*}.

Solution: Using the half angle identity, \begin{align*}\alpha = 30^\circ\end{align*}, and \begin{align*}15^\circ\end{align*} is located in the first quadrant. Therefore, \begin{align*}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*}.

\begin{align*}\sin 15^\circ & = \sqrt{\frac{1 - \cos 30^\circ}{2}} \\ & = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}}\end{align*}

Plugging this into a calculator, \begin{align*}\sqrt{\frac{2 - \sqrt{3}}{4}} \approx 0.2588\end{align*}. Using the sine function on your calculator will validate that this answer is correct.

Example 2: Use the half angle identity to find exact value of \begin{align*}\sin 112.5^\circ\end{align*}

Solution: since \begin{align*}\sin \frac{225^\circ}{2} = \sin 112.5^\circ\end{align*}, use the half angle formula for sine, where \begin{align*}\alpha = 225^\circ\end{align*}. In this example, the angle \begin{align*}112.5^\circ\end{align*} is a second quadrant angle, and the sin of a second quadrant angle is positive.

\begin{align*}\sin 112.5^\circ & = \sin \frac{225^\circ}{2} \\ & = \pm \sqrt{\frac{1 - \cos 225^\circ}{2}} \\ & = + \sqrt{\frac{1 - \left (- \frac{\sqrt{2}}{2} \right )}{2}} \\ & = \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}} \\ & = \sqrt{\frac{2 + \sqrt{2}}{4}}\end{align*}

One of the other formulas that was derived for the cosine of a double angle is:

\begin{align*}\cos 2 \theta = 2 \cos^2 \theta - 1\end{align*}. Set \begin{align*}\theta = \frac{\alpha}{2}\end{align*}, so the equation becomes \begin{align*}\cos 2 \frac{\alpha}{2} = - 1 + 2 \cos^2 \frac{\alpha}{2}\end{align*}. Solving this for \begin{align*}\cos \frac{\alpha}{2}\end{align*}, we get:

\begin{align*} \cos 2 \frac{\alpha}{2} & = 2 \cos^2 \frac{\alpha}{2} - 1 \\ \cos \alpha & = 2 \cos^2 \frac{\alpha}{2} -1 \\ 2 \cos^2 \frac{\alpha}{2} & = 1 + \cos \alpha \\ \cos^2 \frac{\alpha}{2} & = \frac{1 + \cos \alpha}{2} \\ \cos \frac{\alpha}{2} & = \pm \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*}

\begin{align*}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} if \begin{align*}\frac{\alpha}{2}\end{align*} is located in either the first or fourth quadrant.

\begin{align*}\cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} if \begin{align*}\frac{\alpha}{2}\end{align*} is located in either the second or fourth quadrant.

Example 3: Given that the \begin{align*}\cos \theta = \frac{3}{4}\end{align*}, and that \begin{align*}\theta\end{align*} is a fourth quadrant angle, find \begin{align*}\cos \frac{1}{2}\ \theta\end{align*}

Solution: Because \begin{align*}\theta\end{align*} is in the fourth quadrant, the half angle would be in the second quadrant, making the cosine of the half angle negative.

\begin{align*}\cos \frac{\theta}{2} & = - \sqrt{\frac{1 + \cos \theta}{2}} \\ & = - \sqrt{\frac{1 + \frac{3}{4}}{2}} \\ & = - \sqrt{\frac{\frac{7}{4}}{2}} \\ & = - \sqrt{\frac{7}{8}} = - \frac{\sqrt{7}}{2 \sqrt{2}} = - \frac{\sqrt{14}}{4}\end{align*}

Example 4: Use the half angle formula for the cosine function to prove that the following expression is an identity: \begin{align*}2 \cos^2 \frac{x}{2} - \cos x = 1\end{align*}

Solution: Use the formula \begin{align*}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} and substitute it on the left-hand side of the expression.

\begin{align*}2 \left( \sqrt{\frac{1 + \cos \theta}{2}} \right )^2 - \cos \theta & = 1 \\ 2 \left (\frac{1 + \cos \theta}{2} \right ) - \cos \theta & = 1 \\ 1 + \cos \theta - \cos \theta & = 1 \\ 1 & = 1\end{align*}

The half angle identity for the tangent function begins with the reciprocal identity for tangent.

\begin{align*}\tan \alpha = \frac{\sin \alpha}{\cos \alpha} \Rightarrow \tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\end{align*}

The half angle formulas for sine and cosine are then substituted into the identity.

\begin{align*}\tan \frac{\alpha}{2} & = \frac{\sqrt{\frac{1 - \cos \alpha}{2}}}{\sqrt{\frac{1 + \cos \alpha}{2}}} \\ & = \frac{\sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha}} \end{align*}

At this point, you can multiply by either \begin{align*}\frac{\sqrt{1 - \cos \alpha}}{\sqrt{1 - \cos \alpha}}\end{align*} or \begin{align*}\frac{\sqrt{1 + \cos \alpha}}{\sqrt{1 + \cos \alpha}}\end{align*}. We will show both, because they produce different answers.

\begin{align*}& = \frac{\sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha}} \cdot \frac{\sqrt{1 - \cos \alpha}}{\sqrt{1 - \cos \alpha}} && && && = \frac{\sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha}} \cdot \frac{\sqrt{1 + \cos \alpha}}{\sqrt{1 + \cos \alpha}} \\ & = \frac{1 - \cos \alpha}{\sqrt{1 - \cos^2 \alpha}} && \text{or} && && = \frac{\sqrt{1 - \cos^2 \alpha}}{1 + \cos \alpha} \\ & = \frac{1 - \cos \alpha}{\sqrt{\sin^2 \alpha}}&& && && = \frac{\sqrt{\sin^2 \alpha}}{1 + \cos \alpha} \\ & = \frac{1 - \cos \alpha}{\sin \alpha} && && && = \frac{\sin \alpha}{1 + \cos \alpha}\end{align*}

So, the two half angle identities for tangent are \begin{align*}\tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha}\end{align*} and \begin{align*}\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}\end{align*}.

Example 5: Use the half-angle identity for tangent to determine an exact value for \begin{align*}\tan \frac{7 \pi}{12}\end{align*}.

Solution:

\begin{align*}\tan \frac{\alpha}{2} & = \frac{1 - \cos \alpha}{\sin \alpha} \\ \tan \frac{7 \pi}{12} & = \frac{1 - \cos \frac{7 \pi}{6} }{\sin \frac{7 \pi}{6}} \\ \tan \frac{7 \pi}{12} & = \frac{1 + \frac{\sqrt{3}}{2}}{- \frac{1}{2}} \\ \tan \frac{7 \pi}{12} & = - 2 - \sqrt{3}\end{align*}

Example 6: Prove the following identity: \begin{align*}\tan x = \frac{1 - \cos 2x}{\sin 2x}\end{align*}

Solution: Substitute the double angle formulas for \begin{align*}\cos 2x\end{align*} and \begin{align*}\sin 2x\end{align*}.

\begin{align*}\tan x & = \frac{1 - \cos 2x}{\sin 2x} \\ & = \frac{1 - (1 - 2 \sin^2 x)}{2 \sin x \cos x} \\ & = \frac{1 - 1 + 2 \sin^2 x}{2 \sin x \cos x} \\ & = \frac{2 \sin^2 x}{2 \sin x \cos x} \\ & = \frac{\sin x}{\cos x} \\ & = \tan x\end{align*}

Solving Trigonometric Equations Using Half Angle Formulas

Example 7: Solve the trigonometric equation \begin{align*}\sin^2 \theta = 2 \sin^2 \frac{\theta}{2}\end{align*} over the interval \begin{align*}[0, 2\pi)\end{align*}.

Solution:

\begin{align*}\sin^2 \theta & = 2 \sin^2 \frac{\theta}{2} \\ \sin^2 \theta & = 2 \left (\frac{1 - \cos \theta}{2} \right ) && \text{Half angle identity} \\ 1 - \cos^2 \theta & = 1 - \cos \theta && \text{Pythagorean identity} \\ \cos \theta - \cos^2 \theta & = 0 \\ \cos \theta (1 - \cos \theta) & = 0\end{align*}

Then \begin{align*}\cos \theta = 0\end{align*} or \begin{align*}1 - \cos \theta = 0\end{align*}, which is \begin{align*}\cos \theta = 1\end{align*}.

\begin{align*}\theta = 0, \frac{\pi}{2}, \frac{3\pi}{2}, \text{or } 2\pi\end{align*}.

Points to Consider

  • Can you derive a third or fourth angle formula?
  • How do \begin{align*}\frac{1}{2} \sin x\end{align*} and \begin{align*}\sin \frac{1}{2}x\end{align*} differ? Is there a formula for \begin{align*}\frac{1}{2} \sin x\end{align*}?

Review Questions

  1. Find the exact value of:
    1. \begin{align*}\cos 112.5^\circ\end{align*}
    2. \begin{align*}\sin 105^\circ\end{align*}
    3. \begin{align*}\tan \frac{7 \pi}{8}\end{align*}
    4. \begin{align*}\tan \frac{\pi}{8}\end{align*}
    5. \begin{align*}\sin 67.5^\circ\end{align*}
    6. \begin{align*}\tan 165^\circ\end{align*}
  2. If \begin{align*}\sin \theta = \frac{7}{25}\end{align*} and \begin{align*}\theta\end{align*} is in Quad II, find \begin{align*}\sin \frac{\theta}{2}, \cos \frac{\theta}{2}, \tan \frac{\theta}{2}\end{align*}
  3. Prove the identity: \begin{align*}\tan \frac{b}{2} = \frac{\sec b}{\sec b \csc b + \csc b}\end{align*}
  4. Verify the identity: \begin{align*}\cot \frac{c}{2} = \frac{\sin c}{1 - \cos c}\end{align*}
  5. Prove that \begin{align*}\sin x \tan \frac{x}{2} + 2 \cos x = 2 \cos^2 \frac{x}{2}\end{align*}
  6. If \begin{align*}\sin u = - \frac{8}{13}\end{align*}, find \begin{align*}\cos \frac{u}{2}\end{align*}
  7. Solve \begin{align*}2 \cos^2 \frac{x}{2} = 1\end{align*} for \begin{align*}0 \le x < 2 \pi\end{align*}
  8. Solve \begin{align*}\tan \frac{a}{2} = 4\end{align*} for \begin{align*}0^\circ \le a < 360^\circ\end{align*}
  9. Solve the trigonometric equation \begin{align*}\cos \frac{x}{2} = 1 + \cos x\end{align*} such that \begin{align*}0 \le x < 2 \pi\end{align*}.
  10. Prove \begin{align*}\frac{\sin x}{1 + \cos x} = \frac{1 - \cos x}{\sin x}\end{align*}.

Review Answers

    1. \begin{align*}\cos 112.5^\circ & = \cos \frac{225^\circ}{2} = - \sqrt{\frac{1 + \cos 225^\circ}{2}} \\ & = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = - \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} = - \sqrt{\frac{2 - \sqrt{2}}{4}} = - \frac{\sqrt{2 - \sqrt{2}}}{2}\end{align*}
    2. \begin{align*}\sin 105^ \circ & = \sin \frac{210^\circ}{2} = \sqrt{\frac{1 - \cos 210^\circ}{2}} \\ & = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}\end{align*}
    3. \begin{align*} \tan \frac{7 \pi}{8} & = \tan \frac{1}{2} \cdot \frac{7 \pi}{4} = \frac{1 - \cos \frac{7 \pi}{4}}{\sin \frac{7 \pi}{4}} \\ & = \frac{1 - \frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} = \frac{\frac{2 - \sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} = - \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{- 2 \sqrt{2} + 2}{2} = - \sqrt{2} +1\end{align*}
    4. \begin{align*}\tan \frac{\pi}{8} = \tan \frac{1}{2} \cdot \frac{\pi}{4} = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\frac{2 - \sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{2 \sqrt{2} -2}{2} = \sqrt{2} - 1\end{align*}
    5. \begin{align*}\sin 67.5^\circ = \sin \frac{135^\circ}{2} = \sqrt{\frac{1 - \cos 135^\circ}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}\end{align*}
    6. \begin{align*}\tan 165^\circ =\tan \frac{330^\circ}{2} = \frac{1 - \cos 330^\circ}{\sin 330^\circ} = \frac{1 - \frac{\sqrt{3}}{2}}{- \frac{1}{2}} = \frac{\frac{2 - \sqrt{3}}{2}}{- \frac{1}{2}} = - \left (2 - \sqrt{3} \right ) = - 2 + \sqrt{3}\end{align*}
  1. If \begin{align*}\sin \theta = \frac{7}{25}\end{align*}, then by the Pythagorean Theorem the third side is 24. Because \begin{align*}\theta\end{align*} is in the second quadrant, \begin{align*}\cos \theta = - \frac{24}{25}\end{align*}. \begin{align*}\sin \frac{\theta}{2} & = \sqrt{\frac{1 - \cos \theta}{2}} && \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \\ & = \sqrt{\frac{1 + \frac{24}{25}}{2}} && \qquad \ = \sqrt{\frac{1 - \frac{24}{25}}{2}} && \tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \\ & = \sqrt{\frac{49}{50}} && \qquad \ = \sqrt{\frac{1}{50}} && \qquad \ = \sqrt{\frac{1 + \frac{24}{25}}{1 - \frac{24}{25}}} \\ & = \frac{7}{5 \sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}} && \qquad \ = \frac{1}{5 \sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}} && \qquad \ = \sqrt{\frac{49}{50} \cdot \frac{50}{1}} \\ & = \frac{7 \sqrt{2}}{10} && \qquad \ = \frac{\sqrt{2}}{10} && \qquad \ = \sqrt{49} \\ & && && \qquad \ = 7\end{align*}
  2. Step 1: Change right side into sine and cosine. \begin{align*}\tan \frac{b}{2} & = \frac{\sec b}{\sec b \csc b + \csc b} \\ & = \frac{1}{\cos b} \div \csc b (\sec b + 1) \\ & = \frac{1}{\cos b} \div \frac{1}{\sin b} \left (\frac{1}{\cos b} + 1 \right ) \\ & = \frac{1}{\cos b} \div \frac{1}{\sin b} \left (\frac{1 + \cos b}{\cos b} \right ) \\ & = \frac{1}{\cos b} \div \frac{1 + \cos b}{\sin b \cos b} \\ & = \frac{1}{\cos b} \cdot \frac{\sin b \cos b}{1 + \cos b} \\ & = \frac{\sin b}{1 + \cos b}\end{align*} Step 2: At the last step above, we have simplified the right side as much as possible, now we simplify the left side, using the half angle formula. \begin{align*}\sqrt{\frac{1 - \cos b}{1 + \cos b}} & = \frac{\sin b}{1 + \cos b} \\ \frac{1 - \cos b}{1 + \cos b} & = \frac{\sin^2 b}{(1 + \cos b)^2} \\ (1 - \cos b)(1 + \cos b)^2 & = \sin^2 b (1 + \cos b) \\ (1 - \cos b)(1 + \cos b) & = \sin^2 b \\ 1 - \cos^2 b & = \sin^2 b\end{align*}
  3. Step 1: change cotangent to cosine over sine, then cross-multiply. \begin{align*}\cot \frac{c}{2} & = \frac{\sin c}{1 - \cos c} \\ & = \frac{\cos \frac{c}{2}}{\sin \frac{c}{2}} = \sqrt{\frac{1 + \cos c}{1 - \cos c}} \\ \sqrt{\frac{1 + \cos c}{1 - \cos c}} & = \frac{\sin c}{1 - \cos c} \\ \frac{1 + \cos c}{1 - \cos c} & = \frac{\sin^2 c}{(1 - \cos c)^2} \\ (1 + \cos c)(1 - \cos c)^2 & = \sin^2 c (1 - \cos c) \\ (1 + \cos c)(1 - \cos c) & = \sin^2 c \\ 1 - \cos^2 c & = \sin^2 c\end{align*}
  4. \begin{align*}\sin x \tan \frac{x}{2} + 2 \cos x & = \sin x \left(\frac{1 - \cos x}{\sin x} \right ) + 2 \cos x \\ \sin x \tan \frac{x}{2} + 2 \cos x & = 1 - \cos x + 2 \cos x \\ \sin x \tan \frac{x}{2} + 2 \cos x & = 1 + \cos x \\ \sin x \tan \frac{x}{2} + 2 \cos x & = 2 \cos^2 \frac{x}{2}\end{align*}
  5. First, we need to find the third side. Using the Pythagorean Theorem, we find that the final side is \begin{align*}\sqrt{105} \left (b = \sqrt{13^2 - (-8)^2} \right )\end{align*}. Using this information, we find that \begin{align*}\cos u = \pm\frac{\sqrt{105}}{13}\end{align*}. Plugging this into the half angle formula, we get: \begin{align*}\cos \frac{u}{2} & = - \sqrt{\frac{1 \pm \sqrt{\frac{105}{13}}}{2}} \\ & = - \sqrt{\frac{\frac{13 \pm \sqrt{105}}{13}}{2}} \\ & = - \sqrt{\frac{13 \pm \sqrt{105}}{26}}\end{align*}
  6. To solve \begin{align*}2 \cos^2 \frac{x}{2} = 1\end{align*}, first we need to isolate cosine, then use the half angle formula. \begin{align*}2 \cos^2 \frac{x}{2} & = 1 \\ \cos^2 \frac{x}{2} & = \frac{1}{2} \\ \frac{1 + \cos x}{2} & = \frac{1}{2} \\ 1 + \cos x & = 1 \\ \cos x & = 0 \end{align*} \begin{align*}\cos x = 0\end{align*} when \begin{align*} x = \frac{\pi}{2}, \frac{3 \pi}{2}\end{align*}
  7. To solve \begin{align*}\tan \frac{a}{2} = 4\end{align*}, first isolate tangent, then use the half angle formula. \begin{align*}\tan \frac{a}{2} & = 4 \\ \sqrt{\frac{1 - \cos a}{1 + \cos a}} & = 4 \\ \frac{1 - \cos a}{1 + \cos a} & = 16 \\ 16 + 16 \cos a & = 1 - \cos a \\ 17 \cos a & = - 15 \\ \cos a & = - \frac{15}{17}\end{align*} Using your graphing calculator, \begin{align*}\cos a = - \frac{15}{17}\end{align*} when \begin{align*}a = 152^\circ, 208^\circ\end{align*}
  8. \begin{align*}\cos \frac{x}{2} & = 1 + \cos x \\ \pm \sqrt{\frac{1 + \cos x}{2}} & = 1 + \cos x && \text{Half angle identity} \\ \left (\pm \sqrt{\frac{1 + \cos x}{2}} \right )^2 & = (1 + \cos x)^2 && \text{square both sides}\\ \frac{1 + \cos x}{2} & = 1 + 2 \cos x + \cos^2 x \\ 2 \left (\frac{1 + \cos x}{2} \right ) & = 2 (1 + 2 \cos x + \cos^2 x) \\ 1 + \cos x & = 2 + 4 \cos x + 2 \cos^2 x \\ 2 \cos^2 x + 3 \cos x + 1 & = 0 \\ (2 \cos x + 1)(\cos x + 1) & = 0 \\ \text{Then}\quad 2 \cos x + 1 & = 0 \\ \frac{2 \cos x}{2} & = \frac{-1}{2} \\ x & = \frac{2 \pi}{3}, \frac{4 \pi}{3} \\ \text{Or} \quad \cos x + 1 & = 0 \\ \cos x & = - 1 \\ x & = \pi\end{align*}
  9. \begin{align*}\frac{\sin x}{ 1+ \cos x} = \frac{1 - \cos x}{\sin x}\end{align*} This is the two formulas for \begin{align*}\tan \frac{x}{2}\end{align*}. Cross-multiply. \begin{align*}\frac{\sin x}{ 1+ \cos x} & = \frac{1 - \cos x}{\sin x} \\ (1 - \cos x)(1 + \cos x) & = \sin^2 x \\ 1 + \cos x - \cos x - \cos^2 x & = \sin^2 x \\ 1 - \cos^2 x & = \sin^2 x \\ 1 & = \sin^2 x + \cos^2 x\end{align*}

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