📐 Level 2 - Topic 11: Half Angle Identities - Halving the Angle Power ½🔗

1) Introducing Half Angle Identities - Halving the Angle ½

Welcome to the world of half angle identities! Following our exploration of double angle identities, we now venture into identities that express trigonometric functions of \( \frac{\theta}{2} \) in terms of trigonometric functions of \( \theta \). These identities are derived from the double angle identities and are incredibly useful for finding exact values of trigonometric functions for angles that are half of standard angles, and for various manipulations in trigonometry and calculus.

What are Half Angle Identities?

  • They are formulas that express \( \sin(\frac{\theta}{2}) \), \( \cos(\frac{\theta}{2}) \), and \( \tan(\frac{\theta}{2}) \) using trigonometric functions of \( \theta \).
  • They are derived directly from the cosine double angle identities.
  • A key feature is the presence of a \( \pm \) sign in the sine and cosine half-angle identities, determined by the quadrant of \( \frac{\theta}{2} \).

Why are Half Angle Identities Important?

  • Calculating Exact Values: Enable us to find exact trigonometric values for angles like \( 22.5^\circ = \frac{45^\circ}{2} \), \( 15^\circ = \frac{30^\circ}{2} \), \( 11.25^\circ = \frac{22.5^\circ}{2} \), etc.
  • Simplification: Useful in simplifying expressions and integrals in calculus.
  • Deriving Further Identities: They are foundational for further advanced trigonometric identities and techniques.

In this topic, we will derive and utilize the half angle identities for sine, cosine, and tangent. We will learn how to determine the correct sign for sine and cosine identities and explore the multiple forms of the tangent half-angle identity. Let's halve the angle and expand our trigonometric problem-solving toolkit! ½🔗

2) Half Angle Identities for Sine - Sine of Half Angle 🎶

Let's start with the half angle identities for the sine function.

Definition: Half Angle Identities for Sine

The half angle identity for sine is:

\( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \)

Derivation from Cosine Double Angle Identity:
  • Start with the cosine double angle identity (Form 3): \( \cos(2A) = 1 - 2\sin^2(A) \).
  • We want to express sine of a half angle, so let \( A = \frac{\theta}{2} \). Then \( 2A = \theta \). Substitute these into the double angle identity.

    \( \cos(\theta) = 1 - 2\sin^2\left(\frac{\theta}{2}\right) \)

  • Solve for \( \sin^2\left(\frac{\theta}{2}\right) \):

    \( 2\sin^2\left(\frac{\theta}{2}\right) = 1 - \cos(\theta) \)

    \( \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{2} \)

  • Take square root of both sides:

    \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \)

  • Sign Determination: The sign \( \pm \) is determined by the quadrant in which \( \frac{\theta}{2} \) lies. Sine is positive in quadrants I and II, and negative in quadrants III and IV.

Example 1: Using Sine Half Angle Identity

Find the exact value of \( \sin(15^\circ) \) using the half angle identity.

  1. Recognize \( 15^\circ \) as half of a known angle: \( 15^\circ = \frac{30^\circ}{2} \). So, \( \frac{\theta}{2} = 15^\circ \Rightarrow \theta = 30^\circ \).
  2. Determine the sign: \( 15^\circ \) is in quadrant I, where sine is positive. So, we use the positive square root.
  3. Apply sine half angle identity: \( \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{2}} \) with \( \theta = 30^\circ \).

    \( \sin(15^\circ) = \sin\left(\frac{30^\circ}{2}\right) = \sqrt{\frac{1 - \cos(30^\circ)}{2}} \)

  4. Substitute known value: \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).

    \( \sin(15^\circ) = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} \)

  5. Simplify:

    \( \sin(15^\circ) = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2} \)

Solution: \( \sin(15^\circ) = \frac{\sqrt{2 - \sqrt{3}}}{2} \). (Note: This form looks different from \( \frac{\sqrt{6} - \sqrt{2}}{4} \) obtained using the difference identity in Topic 9, but they are equivalent! Verifying this equivalence is an interesting exercise!).

3) Half Angle Identities for Cosine - Cosine of Half Angle 🎵

Now, let's derive and explore the half angle identities for the cosine function.

Definition: Half Angle Identities for Cosine

The half angle identity for cosine is:

\( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}} \)

Derivation from Cosine Double Angle Identity:
  • Start with the cosine double angle identity (Form 2): \( \cos(2A) = 2\cos^2(A) - 1 \).
  • Let \( A = \frac{\theta}{2} \). Then \( 2A = \theta \). Substitute these into the double angle identity.

    \( \cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right) - 1 \)

  • Solve for \( \cos^2\left(\frac{\theta}{2}\right) \):

    \( 2\cos^2\left(\frac{\theta}{2}\right) = 1 + \cos(\theta) \)

    \( \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \cos(\theta)}{2} \)

  • Take square root of both sides:

    \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}} \)

  • Sign Determination: The sign \( \pm \) is determined by the quadrant in which \( \frac{\theta}{2} \) lies. Cosine is positive in quadrants I and IV, and negative in quadrants II and III.

Example 2: Using Cosine Half Angle Identity

Find the exact value of \( \cos(22.5^\circ) \) using the half angle identity.

  1. Recognize \( 22.5^\circ \) as half of a known angle: \( 22.5^\circ = \frac{45^\circ}{2} \). So, \( \frac{\theta}{2} = 22.5^\circ \Rightarrow \theta = 45^\circ \).
  2. Determine the sign: \( 22.5^\circ \) is in quadrant I, where cosine is positive. So, we use the positive square root.
  3. Apply cosine half angle identity: \( \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}} \) with \( \theta = 45^\circ \).

    \( \cos(22.5^\circ) = \cos\left(\frac{45^\circ}{2}\right) = \sqrt{\frac{1 + \cos(45^\circ)}{2}} \)

  4. Substitute known value: \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \).

    \( \cos(22.5^\circ) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} \)

  5. Simplify:

    \( \cos(22.5^\circ) = \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \)

Solution: \( \cos(22.5^\circ) = \frac{\sqrt{2 + \sqrt{2}}}{2} \).

4) Half Angle Identities for Tangent - Tangent of Half Angle 🥁

Finally, let's explore the half angle identities for the tangent function. Tangent has multiple useful forms for its half angle identity.

Definition: Half Angle Identities for Tangent

There are three common forms for the half angle identity for tangent:

Form 1 (using square root, sign determined by quadrant of \( \frac{\theta}{2} \)):

\( \tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \)

Form 2 (in terms of sine and cosine, no sign ambiguity, preferred for most simplification):

\( \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \)

Form 3 (another form in terms of sine and cosine, no sign ambiguity, also useful):

\( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \)

Derivation and Relationships:
  • Form 1 Derivation (from sine and cosine half angle):

    \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \frac{\pm \sqrt{\frac{1 - \cos(\theta)}{2}}}{\pm \sqrt{\frac{1 + \cos(\theta)}{2}}} = \pm \sqrt{\frac{\frac{1 - \cos(\theta)}{2}}{\frac{1 + \cos(\theta)}{2}}} = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \)

    The sign \( \pm \) is determined by the quadrant of \( \frac{\theta}{2} \) (tangent is positive in quadrants I and III, negative in II and IV).
  • Form 2 Derivation (multiply Form 1 by \( \sqrt{1 + \cos(\theta)} / \sqrt{1 + \cos(\theta)} \) and simplify):

    \( \tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \cdot \frac{\sqrt{1 + \cos(\theta)}}{\sqrt{1 + \cos(\theta)}} = \pm \frac{\sqrt{1 - \cos^2(\theta)}}{1 + \cos(\theta)} = \pm \frac{\sqrt{\sin^2(\theta)}}{1 + \cos(\theta)} = \pm \frac{|\sin(\theta)|}{1 + \cos(\theta)} \)

    If we consider the sign properties more carefully, and consider the case where we aim for a form without \( \pm \) ambiguity, we can further manipulate.
    Considering sign conventions more carefully, and choosing the positive root for \( \sqrt{\sin^2(\theta)} = |\sin(\theta)| \) under appropriate quadrant considerations to remove ambiguity, we get:

    \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \)

  • Form 3 Derivation (multiply Form 2 by \( (1 - \cos(\theta)) / (1 - \cos(\theta)) \) and simplify):

    \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \cdot \frac{1 - \cos(\theta)}{1 - \cos(\theta)} = \frac{\sin(\theta)(1 - \cos(\theta))}{1 - \cos^2(\theta)} = \frac{\sin(\theta)(1 - \cos(\theta))}{\sin^2(\theta)} \)

    \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \)

  • Relationship between Form 2 and Form 3: Both Form 2 and Form 3 are equivalent and do not have the \( \pm \) sign ambiguity of Form 1. They are often more convenient for simplification and calculations.

Example 3: Using Tangent Half Angle Identity (Form 1)

Find the exact value of \( \tan(22.5^\circ) \) using Form 1 of the half angle identity.

  1. Recognize \( 22.5^\circ \) as half of \( 45^\circ \): \( 22.5^\circ = \frac{45^\circ}{2} \). So, \( \frac{\theta}{2} = 22.5^\circ \Rightarrow \theta = 45^\circ \).
  2. Determine the sign: \( 22.5^\circ \) is in quadrant I, where tangent is positive. So, use the positive square root.
  3. Apply tangent half angle identity (Form 1): \( \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \) with \( \theta = 45^\circ \).

    \( \tan(22.5^\circ) = \tan\left(\frac{45^\circ}{2}\right) = \sqrt{\frac{1 - \cos(45^\circ)}{1 + \cos(45^\circ)}} \)

  4. Substitute known value: \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \).

    \( \tan(22.5^\circ) = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}} = \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}}} = \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} \)

  5. Rationalize denominator inside the square root: Multiply numerator and denominator by \( (2 - \sqrt{2}) \).

    \( \tan(22.5^\circ) = \sqrt{\frac{(2 - \sqrt{2})}{(2 + \sqrt{2})} \cdot \frac{(2 - \sqrt{2})}{(2 - \sqrt{2})}} = \sqrt{\frac{(2 - \sqrt{2})^2}{2^2 - (\sqrt{2})^2}} = \sqrt{\frac{(2 - \sqrt{2})^2}{4 - 2}} = \sqrt{\frac{(2 - \sqrt{2})^2}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} \)

  6. Further simplify (optional, rationalize denominator again): Multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \).

    \( \tan(22.5^\circ) = \frac{2 - \sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1 \)

Solution: \( \tan(22.5^\circ) = \sqrt{2} - 1 \).

Example 4: Using Tangent Half Angle Identity (Form 2)

Find the exact value of \( \tan(15^\circ) \) using Form 2 of the half angle identity: \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \).

  1. Recognize \( 15^\circ \) as half of \( 30^\circ \): \( 15^\circ = \frac{30^\circ}{2} \). So, \( \frac{\theta}{2} = 15^\circ \Rightarrow \theta = 30^\circ \).
  2. Apply tangent half angle identity (Form 2): \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \) with \( \theta = 30^\circ \).

    \( \tan(15^\circ) = \tan\left(\frac{30^\circ}{2}\right) = \frac{\sin(30^\circ)}{1 + \cos(30^\circ)} \)

  3. Substitute known values: \( \sin(30^\circ) = \frac{1}{2} \), \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).

    \( \tan(15^\circ) = \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = \frac{\frac{1}{2}}{\frac{2 + \sqrt{3}}{2}} = \frac{1}{2 + \sqrt{3}} \)

  4. Rationalize denominator: Multiply by \( \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \).

    \( \tan(15^\circ) = \frac{1}{(2 + \sqrt{3})} \cdot \frac{(2 - \sqrt{3})}{(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \)

Solution: \( \tan(15^\circ) = 2 - \sqrt{3} \). (Consistent with the result from tangent difference identity in Topic 9! Form 2 is often easier to work with than Form 1 due to no square root in denominator after substitution).

Example 5: Using Tangent Half Angle Identity (Form 3)

Find the exact value of \( \tan(15^\circ) \) using Form 3 of the half angle identity: \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \).

  1. Recognize \( 15^\circ \) as half of \( 30^\circ \): \( 15^\circ = \frac{30^\circ}{2} \). So, \( \frac{\theta}{2} = 15^\circ \Rightarrow \theta = 30^\circ \).
  2. Apply tangent half angle identity (Form 3): \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \) with \( \theta = 30^\circ \).

    \( \tan(15^\circ) = \tan\left(\frac{30^\circ}{2}\right) = \frac{1 - \cos(30^\circ)}{\sin(30^\circ)} \)

  3. Substitute known values: \( \sin(30^\circ) = \frac{1}{2} \), \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).

    \( \tan(15^\circ) = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3} \)

Solution: \( \tan(15^\circ) = 2 - \sqrt{3} \). (Again, consistent result! Form 3 is also very convenient for calculation).

5) Verifying Identities using Half Angle Identities 🚀

Example 6: Verifying Identity using Sine Half Angle Identity

Verify the identity: \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \) starting from \( \tan\left(\frac{\theta}{2}\right) = \sin\left(\frac{\theta}{2}\right) / \cos\left(\frac{\theta}{2}\right) \) and using half-angle identities for sine and cosine.

  1. Start with the left side (LS): \( \text{LS} = \tan\left(\frac{\theta}{2}\right) = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \).
  2. Apply sine and cosine half angle identities (with appropriate signs assuming quadrant I for simplicity of verification):

    \( \text{LS} = \frac{\sqrt{\frac{1 - \cos(\theta)}{2}}}{\sqrt{\frac{1 + \cos(\theta)}{2}}} \)

  3. Simplify the fraction:

    \( \text{LS} = \sqrt{\frac{\frac{1 - \cos(\theta)}{2}}{\frac{1 + \cos(\theta)}{2}}} = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \)

  4. Multiply inside the square root by \( \frac{1 - \cos(\theta)}{1 - \cos(\theta)} \) to manipulate towards desired form:

    \( \text{LS} = \sqrt{\frac{(1 - \cos(\theta))}{(1 + \cos(\theta))} \cdot \frac{(1 - \cos(\theta))}{(1 - \cos(\theta))}} = \sqrt{\frac{(1 - \cos(\theta))^2}{1 - \cos^2(\theta)}} = \sqrt{\frac{(1 - \cos(\theta))^2}{\sin^2(\theta)}} \)

  5. Simplify the square root (assuming \( \sin(\theta) > 0 \) and \( 1 - \cos(\theta) > 0 \) for simplicity):

    \( \text{LS} = \frac{1 - \cos(\theta)}{\sin(\theta)} \)

  6. Compare with the right side (RS): \( \text{RS} = \frac{1 - \cos(\theta)}{\sin(\theta)} \).
  7. Conclusion: Since \( \text{LS} = \text{RS} \), the identity \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \) is verified.

6) Solving Equations using Half Angle Identities 🚀

Example 7: Solving Equation using Cosine Half Angle Identity (Conceptual – Illustrative Example, equations directly solvable by half-angle identities are less common in basic contexts)

Solve the equation (for illustrative purposes, typical equations are not directly of this form but transformations might lead to it): \( \cos\left(\frac{x}{2}\right) = \cos(x) \) for principal solutions in \( [0, 2\pi) \) (This is more to show how half-angle identities *could* be used in principle, rather than typical application).

  1. Apply cosine half angle identity: Replace \( \cos\left(\frac{x}{2}\right) \) with \( \pm \sqrt{\frac{1 + \cos(x)}{2}} \). Equation becomes \( \pm \sqrt{\frac{1 + \cos(x)}{2}} = \cos(x) \).
  2. Square both sides (careful about introducing extraneous solutions): \( \frac{1 + \cos(x)}{2} = \cos^2(x) \).
  3. Rearrange into a quadratic equation in \( \cos(x) \): \( 1 + \cos(x) = 2\cos^2(x) \Rightarrow 2\cos^2(x) - \cos(x) - 1 = 0 \).
  4. Solve quadratic equation: Factor \( (2\cos(x) + 1)(\cos(x) - 1) = 0 \). Solutions are \( \cos(x) = -\frac{1}{2} \) or \( \cos(x) = 1 \).
  5. Solve \( \cos(x) = -\frac{1}{2} \): Principal solutions in \( [0, 2\pi) \) are \( x = \frac{2\pi}{3}, \frac{4\pi}{3} \).
  6. Solve \( \cos(x) = 1 \): Principal solution in \( [0, 2\pi) \) is \( x = 0 \).
  7. Check for extraneous solutions (due to squaring) in the original equation \( \cos\left(\frac{x}{2}\right) = \cos(x) \):
    • For \( x = 0 \): \( \cos(0/2) = \cos(0) = 1 \). Valid.
    • For \( x = \frac{2\pi}{3} \): \( \cos(\frac{\pi}{3}) = \frac{1}{2} \). \( \cos(\frac{2\pi}{3}) = -\frac{1}{2} \). Not valid. Extraneous.
    • For \( x = \frac{4\pi}{3} \): \( \cos(\frac{2\pi}{3}) = -\frac{1}{2} \). \( \cos(\frac{4\pi}{3}) = -\frac{1}{2} \). Valid.
  8. Principal solutions: \( x = 0, \frac{4\pi}{3} \). Note: We needed to check for extraneous solutions because of squaring. In typical use cases, half-angle identities are often used to *simplify* before solving, rather than directly in equation solving as shown here for illustration.

Solution: Principal solutions are \( x = 0, \frac{4\pi}{3} \).

7) Practice Questions 🎯

7.1 Fundamental – Using Half Angle Identities for Calculation and Verification

1. Find the exact value of \( \sin(22.5^\circ) \).

2. Find the exact value of \( \cos(11.25^\circ) \) (Hint: \( 11.25^\circ = \frac{22.5^\circ}{2} \)).

3. Find the exact value of \( \tan(22.5^\circ) \) using Form 2: \( \tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1 + \cos(\theta)} \).

4. Find the exact value of \( \tan(22.5^\circ) \) using Form 3: \( \tan(\frac{\theta}{2}) = \frac{1 - \cos(\theta)}{\sin(\theta)} \).

5. Find the exact value of \( \sin(195^\circ) \) (Hint: \( 195^\circ = \frac{390^\circ}{2} = \frac{360^\circ + 30^\circ}{2} = 180^\circ + 15^\circ \), consider quadrant for sign).

6. Verify the identity: \( \cos(\frac{\theta}{2}) = \frac{\sin(\theta)}{2\sin(\frac{\theta}{2})} \).

7. Verify the identity: \( \sin(\theta) = \pm \sqrt{\frac{1 - \cos(2\theta)}{2}} \) by replacing \( \theta \) with \( 2\theta \) in the sine half-angle identity and adjusting notation.

8. Verify the identity: \( \cos(\theta) = \pm \sqrt{\frac{1 + \cos(2\theta)}{2}} \) by similar substitution as in problem 7.

9. Simplify: \( \sqrt{\frac{1 - \cos(4x)}{2}} \) assuming \( \sin(2x) \ge 0 \).

10. Simplify: \( \sqrt{\frac{1 + \cos(6x)}{2}} \) assuming \( \cos(3x) \ge 0 \).

11. Simplify: \( \frac{\sin(x)}{1 + \cos(x)} \).

12. Simplify: \( \frac{1 - \cos(x)}{\sin(x)} \).

7.2 Challenging – Advanced Verification and Equation Solving 💪🚀

1. Verify: \( \frac{1 - \cos(\theta)}{1 + \cos(\theta)} = \left(\frac{\sin(\theta)}{1 + \cos(\theta)}\right)^2 \).

2. Verify: \( \frac{1 + \cos(\theta)}{\sin(\theta)} + \frac{\sin(\theta)}{1 + \cos(\theta)} = 2\csc(\theta) \).

3. Verify: \( \tan\left(\frac{\theta}{2}\right) = \csc(\theta) - \cot(\theta) \) using half angle identity for tangent (Form 3).

4. Solve the equation: \( 2\cos^2\left(\frac{x}{2}\right) - 3\cos(x) = 0 \) for principal solutions in \( [0, 2\pi) \).

5. Solve the equation: \( \sin\left(\frac{x}{2}\right) + \cos(x) = 1 \) for principal solutions in \( [0^\circ, 360^\circ) \) (Consider sign carefully if using square root form).

6. (Conceptual) Explain why there is a \( \pm \) sign in the half-angle identities for sine and cosine but not in Form 2 and Form 3 for tangent. When is it crucial to consider the \( \pm \) sign?

7. (Challenging Verification & Simplification) Simplify \( \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} + \frac{1 - \cos(\theta)}{\sin(\theta)} \) and show it is equal to \( \frac{1}{\sin(\theta)} \) for \( 0 < \theta < \pi \).

8) Summary - Half Angle Identities - Power of Halving 🎉

  • Half Angle Identities: Formulas for \( \sin(\frac{\theta}{2}) \), \( \cos(\frac{\theta}{2}) \), \( \tan(\frac{\theta}{2}) \) in terms of functions of \( \theta \).
  • Sine Half Angle: \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \) (sign depends on quadrant of \( \frac{\theta}{2} \)).
  • Cosine Half Angle: \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}} \) (sign depends on quadrant of \( \frac{\theta}{2} \)).
  • Tangent Half Angle: Three forms:
    • Form 1: \( \tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \) (sign depends on quadrant of \( \frac{\theta}{2} \)).
    • Form 2: \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \) (no sign ambiguity).
    • Form 3: \( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \) (no sign ambiguity).
  • Derivation: Derived from cosine double angle identities.
  • Applications: Calculating exact values of trigonometric functions for half angles, simplifying expressions, and further trigonometric manipulations.
  • Techniques: Recognize half angle forms, determine the correct sign when using \( \pm \) form, choose the appropriate tangent half-angle form based on the context, algebraic simplification, careful checking for extraneous solutions when solving equations.

Congratulations! You have now added half angle identities to your trigonometric toolkit. These identities not only expand the range of angles for which you can find exact trigonometric values but also provide powerful simplification tools for more advanced trigonometric problems. Understanding the nuances of sign determination and choosing the right form of the tangent half-angle identity are key skills you've developed in this topic. As you continue your trigonometric journey, you'll find these half-angle identities to be invaluable. Keep practicing, and you'll master the art of halving angles to solve even more fascinating problems! ½🔗💪📐🌟

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