How to Solve: Half Angle Identities

How to Solve: Half Angle Identities

Introduction

"Half-angle identities let you find exact values of sin(15°), cos(22.5°), or even prove trigonometric identities on your exam—without a calculator. Master these, and you’ll solve problems faster than your classmates."

What You Need To Know First

1. Pythagorean Identities – You must know:
2. ( \sin^2 \theta + \cos^2 \theta = 1 )
3. ( 1 + \tan^2 \theta = \sec^2 \theta )

4. ( 1 + \cot^2 \theta = \csc^2 \theta )

5. Double Angle Formulas – You must recall:

6. ( \cos(2\theta) = 1 - 2\sin^2 \theta )
7. ( \cos(2\theta) = 2\cos^2 \theta - 1 )

8. ( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta )

9. Algebraic Manipulation – You must be comfortable solving for ( \sin \theta ) or ( \cos \theta ) in equations.

Key Vocabulary

Formulas To Know

1. Sine Half-Angle Identity

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

Variables: - ( \theta ) = any angle (in degrees or radians) - The ( \pm ) sign depends on the quadrant of ( \theta/2 ).

MEMORISE THIS – Not always given on exam sheets.

2. Cosine Half-Angle Identity

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

Variables: - ( \theta ) = any angle - The ( \pm ) sign depends on the quadrant of ( \theta/2 ).

MEMORISE THIS – Not always given on exam sheets.

3. Tangent Half-Angle Identities

Formula (Option 1): [ \tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} ]

Formula (Option 2): [ \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{\sin \theta} ]

Formula (Option 3): [ \tan\left(\frac{\theta}{2}\right) = \frac{\sin \theta}{1 + \cos \theta} ]

Variables: - ( \theta ) = any angle - The ( \pm ) sign depends on the quadrant of ( \theta/2 ).

GIVEN ON EXAM SHEET (usually) – But memorising Option 2 or 3 saves time.

Step-by-Step Method

Step 1: Identify the Angle

• Write down the angle you need to find (e.g., ( \sin(15°) )).
• Express it as half of a known angle (e.g., ( 15° = 30°/2 )).

Step 2: Choose the Correct Half-Angle Formula

• If finding sine, use ( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} ).
• If finding cosine, use ( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} ).
• If finding tangent, use one of the three tangent formulas (Option 2 or 3 is easiest).

Step 3: Plug in the Known Angle

• Substitute ( \theta ) into the formula (e.g., ( \theta = 30° ) for ( \sin(15°) )).

Step 4: Compute ( \cos \theta ) or ( \sin \theta )

• Find the exact value of ( \cos \theta ) or ( \sin \theta ) (e.g., ( \cos(30°) = \frac{\sqrt{3}}{2} )).

Step 5: Simplify Inside the Square Root

• Substitute and simplify the expression inside the square root (e.g., ( \frac{1 - \frac{\sqrt{3}}{2}}{2} )).

Step 6: Determine the Sign (( \pm ))

• Find the quadrant of ( \theta/2 ).
• Use the ASTC (All Students Take Calculus) rule:
• A (Quadrant I) → All positive
• S (Quadrant II) → Sine positive
• T (Quadrant III) → Tangent positive
• C (Quadrant IV) → Cosine positive
• Choose the correct sign based on the quadrant.

Step 7: Write the Final Answer

• Simplify the expression and write the exact value.

Worked Example Using the Steps

Problem: Find the exact value of ( \cos(22.5°) ).

Step 1: Identify the angle. - ( 22.5° = \frac{45°}{2} ), so ( \theta = 45° ).

Step 2: Choose the cosine half-angle formula. - ( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} ).

Step 3: Plug in ( \theta = 45° ). - ( \cos(22.5°) = \pm \sqrt{\frac{1 + \cos(45°)}{2}} ).

Step 4: Compute ( \cos(45°) ). - ( \cos(45°) = \frac{\sqrt{2}}{2} ).

Step 5: Simplify inside the square root. - ( \frac{1 + \frac{\sqrt{2}}{2}}{2} = \frac{\frac{2 + \sqrt{2}}{2}}{2} = \frac{2 + \sqrt{2}}{4} ).

Step 6: Determine the sign. - ( 22.5° ) is in Quadrant I → cosine is positive. - So, ( \cos(22.5°) = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} ).

Step 7: Final answer. - ( \cos(22.5°) = \frac{\sqrt{2 + \sqrt{2}}}{2} ).

Worked Examples

Example 1 - Basic

Problem: Find ( \sin(15°) ).

Solution: 1. ( 15° = \frac{30°}{2} ), so ( \theta = 30° ). 2. Use the sine half-angle formula:
[ \sin(15°) = \pm \sqrt{\frac{1 - \cos(30°)}{2}} ] 3. ( \cos(30°) = \frac{\sqrt{3}}{2} ). 4. Substitute:
[ \sin(15°) = \pm \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \pm \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \pm \sqrt{\frac{2 - \sqrt{3}}{4}} ] 5. ( 15° ) is in Quadrant I → sine is positive. 6. Final answer:
[ \sin(15°) = \frac{\sqrt{2 - \sqrt{3}}}{2} ]

What we did and why: - We expressed ( 15° ) as half of ( 30° ) to use the half-angle formula. - We simplified the expression inside the square root and chose the correct sign based on the quadrant.

Example 2 - Medium

Problem: Find ( \tan(105°) ) using a half-angle identity.

Solution: 1. ( 105° = \frac{210°}{2} ), so ( \theta = 210° ). 2. Use the tangent half-angle formula (Option 2):
[ \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{\sin \theta} ] 3. ( \cos(210°) = -\frac{\sqrt{3}}{2} ), ( \sin(210°) = -\frac{1}{2} ). 4. Substitute:
[ \tan(105°) = \frac{1 - \left(-\frac{\sqrt{3}}{2}\right)}{-\frac{1}{2}} = \frac{1 + \frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\left(2 + \sqrt{3}\right) ] 5. Final answer:
[ \tan(105°) = -2 - \sqrt{3} ]

What we did and why: - We chose ( \theta = 210° ) because ( 105° ) is half of it. - We used the tangent half-angle formula that avoids square roots for simplicity. - We simplified the fraction and kept the negative sign because ( 105° ) is in Quadrant II, where tangent is negative.

Example 3 - Exam Style

Problem: Prove the identity: [ \frac{1 - \cos(2x)}{\sin(2x)} = \tan(x) ]

Solution: 1. Recognise the left side resembles a half-angle formula. 2. Recall the tangent half-angle identity:
[ \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{\sin \theta} ] 3. Let ( \theta = 2x ), so:
[ \tan\left(\frac{2x}{2}\right) = \frac{1 - \cos(2x)}{\sin(2x)} ] 4. Simplify:
[ \tan(x) = \frac{1 - \cos(2x)}{\sin(2x)} ] 5. This matches the left side of the original equation, so the identity is proven.

What we did and why: - We spotted that the left side matched a known half-angle identity. - We substituted ( \theta = 2x ) to rewrite the expression in terms of ( \tan(x) ). - This is a common exam trick—recognising disguised identities.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in for your exam. Half-angle identities let you find exact values for angles like 15°, 22.5°, or even prove identities. Here’s the game plan:

1. Memorise the formulas:
2. ( \sin(\theta/2) = \pm \sqrt{\frac{1 - \cos \theta}{2}} )
3. ( \cos(\theta/2) = \pm \sqrt{\frac{1 + \cos \theta}{2}} )

4. ( \tan(\theta/2) = \frac{1 - \cos \theta}{\sin \theta} ) (easiest one to use).

5. Always check the quadrant of ( \theta/2 ) to pick the right sign. Quadrant I? All positive. Quadrant II? Sine positive, cosine negative. And so on.

6. For proofs, look for expressions like ( \frac{1 - \cos(2x)}{\sin(2x)} )—that’s just ( \tan(x) ) in disguise.

7. Practice the steps:

8. Identify ( \theta ).
9. Plug into the formula.
10. Simplify.
11. Pick the sign.
12. Write the exact value.

Tonight, try finding ( \sin(75°) ) and ( \cos(165°) ) using half-angle identities. If you can do those, you’re ready for the exam. Good luck!