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Half-Angle Formulas

Main formulas:

sin 1
2
 x
=
±    /
√
1
2
 (1− cosx)
 
cos 1
2
 x
=
±    /
√
1
2
 (1+cosx)
 

Example 1:

Use the half-angle formulas to find the sine and cosine of 15° . Check that the answers satisfy the Pythagorean identity sin2 x + cos2 x = 1.

Example 2:

Prove the following trigonometric identity:
cos 1
2
 x + sin 1
2
 x
cos 1
2
 x − sin 1
2
 x
 = secx + tanx

Example 3:

Prove the following half-angle formula for tangents. Be careful about removing any ± signs!
tan 1
2
 x = 1− cosx
sinx

Example 4:

Use the half-angle formulas to find the sine and cosine of (π /8). Check that the answers satisfy the Pythagorean identity sin2 x + cos2 x = 1.

Example 5:

Prove the following trigonometric identity:
tanx tan 1
2
 x = secx − 1

Use the half - angle formula to find the sine of [(5π)/8]

  • Half - angle Formula: sin[1/2]x = ±√{[1/2](1 − cosx)}
  • [1/2]x = [(5π)/8] x = [(5π)/4]
  • sin[(5π)/8] = ±√{[1/2](1 − cos[(5π)/4])}
  • sin[(5π)/8] = ±√{[1/2](1 − ( − [(√2 )/2])} ) = ±√{[1/2]([(2 + √2 )/2])} = ±√{[(2 + √2 )/4]} = ±[(√{2 + √2 } )/2]
  • sin[(5π)/8] will be positive because [(5π)/8] is in quadrant II

sin[(5π)/8] = [(√{2 + √2 } )/2]

Use the half - angle formula to find the cosine of [(5π)/8]

  • Half - angle Formula: cos[1/2]x = ±√{[1/2](1 + cosx)}
  • [1/2]x = [5p/8] x = [5p/4]
  • cos[(5π)/8] = ±√{[1/2](1 + cos[5p/4])}
  • cos[(5π)/8] = ±√{[1/2](1 + ( − [(√2 )/2])} ) = ±√{[1/2]([(2 − √2 )/2])} = ±√{[(2 − √2 )/4]} = ±[(√{2 − √2 } )/2]
  • cos[(5π)/8] will be negative because [(5π)/8] is in quadrant II

cos[(5π)/8] = − [(√{2 − √2 } )/2]

Use the half - angle formula to find the sine of 75°

  • Half - angle Formula: sin[1/2]x = ±√{[1/2](1 − cosx)}
  • [1/2]x = 75° ⇒ x = 150°
  • sin75° = ±√{[1/2](1 − cos150°)}
  • sin75° = ±√{[1/2](1 − ( − [(√3 )/2])} ) = ±√{[1/2]([(2 + √3 )/2])} = ±√{[(2 + √3 )/4]} = ±[(√{2 + √3 } )/2]
  • sin75° will be positive because 75° is in quadrant I

sin75° = [(√{2 + √3})/2]

Use the half - angle formula to find the cosine of 75°

  • Half - angle Formula: cos[1/2]x = ±√{[1/2](1 + cosx)}
  • [1/2]x = 75° ⇒ x = 150°
  • cos75° = ±√{[1/2](1 + cos150°)}
  • cos75° = ±√{[1/2](1 + ( − [(√3 )/2])} ) = ±√{[1/2]([(2 − √3 )/2])} = ±√{[(2 − √3 )/4]} = ±[(√{2 − √3 } )/2]
  • cos75° will be positive because 75° is in quadrant I

cos75° = [(√{2 − √3 } )/2]

Use the half - angle formula to find the sine of 180°

  • Half - angle Formula: sin[1/2]x = ±√{[1/2](1 − cosx)}
  • [1/2]x = 180° ⇒ x = 360°
  • sin180° = ±√{[1/2](1 − cos360°)}
  • sin180° = ±√{[1/2](1 − 1} ) = ±√{[1/2](0)} = 0

sin180° = 0

Use the half - angle formula to find the cosine of 75°

  • Half - angle Formula: cos[1/2]x = ±√{[1/2](1 + cosx)}
  • [1/2]x = 180° ⇒ x = 360°
  • cos180° = ±√{[1/2](1 + cos360°)}
  • cos180° = ±√{[1/2](1 + 1} ) = ±√{[1/2](2)} = ±√1 = ±1

cos180° = 1

Use the half - angle formula to simplify √{[(1 − cos6x)/2]}

  • √{[(1 − cos6x)/2]} · [([1/2])/([1/2])]

|sin3x|

Use the half - angle formula to simplify √{[(1 + cos4x)/2]}

  • √{[(1 + cos4x)/2]} · [([1/2])/([1/2])]

|cos2x|

Use the half - angle formula to simplify − √{[(1 − cos8x)/(1 + cos8x)]}

  • − √{[(1 − cos8x)/(1 + cos8x)]} · [([1/2])/([1/2])]
  • − |[sin4x/cos4x]|

- |tan4x|

Verify the following identity csc2θ = [(cscθ)/(2cosθ)]

  • [1/(sin2θ)] = [(cscθ)/(2cosθ)], Reciprocal Identity
  • [1/(2sinθcosθ)] = [(cscθ)/(2cosθ)], Half - angle Identity
  • [1/(sinθ)] ·[1/(2cosθ)] = [(cscθ)/(2cosθ)], Separate fractions

[(cscθ)/(2cosθ)] = [(cscθ)/(2cosθ)]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Half-Angle Formulas

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Mathematics: Trigonometry

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