Half Angle Formulae
Trig Frog presents some formulae that are easily derived from the Power Reducing Formulae:
sin A/2 = ± £((1 - cosA)/2)
cos A/2 = ± £((1 + cosA)/2)
tan A/2 = ± £((1 - cosA)/(1 + cosA))
These formulae can be proven by replacing the A in the Power Reducing Formulae with A/2. Thus, they become the Half Angle Formulae. Observe:
sin^2 A = (1 - cos2A)/2
sin^2 A/2 = (1 - cos2(A/2))/2 Make the substitution.
= (1 - cos A)/2 Simplify, then take the square root of both sides.
Answer: sin A/2 = ± £((1 - cosA)/2)
cos^2 A = (1 + cos2A)/2
cos^2 A/2 = (1 + cos2(A/2))/2 Make the substitution.
= (1 + cosA)/2 Simplify, then take the square root of both sides.
Answer: cos A/2 = ± £((1 + cosA)/2)
tan^2 A = (1 - cos2A)/(1 + cos2A)
tan^2 A/2 = (1 - cos2(A/2))/(1 + cos2(A/2)) Make the substitution.
= (1 - cosA)/(1 + cosA) Simplify, then take the square root of both sides.
Answer: tan A/2 = ± £((1 - cosA)/(1 + cosA))
Note: The tangent function can also be simplified by rationalizing either the denominator or the numerator to obtain tan A/2 = sinA/(1 + cosA) and tan A/2 = (1 - cosA)/sinA
Example
Find the exact value of sin105° without using the Sum and Difference Formulae.
Solution:
Trig Frog reminds you that the angle 105° is half of the angle 210°. Thus, thou canst use the Half Angle Formulae. And remember, the angle 105° lies in Quadrant II.
sin 105° = £((1 - cos 210°)/2) Power Reduction Formulae.
=£((1 - (-cos 30°))/2) The £ is positive because sine is positive in Quadrant II.
=£((1 + £3/2)/2) Obtain these function values from the Unit Circle.
=£(2+£3)/2 Perform these operations and simplify.
Answer: sin 105° = (£(2+£3))/2
