cos θ = 7/25 and θ is acute. Find sin(θ/2).
- Use sin2(θ/2) = (1 − cos θ)/2 = (1 − 7/25)/2.
- 1 − 7/25 = 18/25, and 18/25 ÷ 2 = 9/25.
- θ/2 is acute, so take the positive root: sin(θ/2) = √(9/25) = 3/5.
Answer: sin(θ/2) = 3/5.
Double, half and triple angles; sums and products
Maths · Trigonometric identities. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
Put β = α in the compound angle formulas and you get the double-angle identities. They give sin 2θ, cos 2θ and tan 2θ from the ratios of θ. Rearranging cos 2θ gives the half-angle identities, and using the formulas once more gives the triple-angle identities.
cos 2θ has three forms. Choose the one that fits the information you have: only sin θ, only cos θ, or both.
Adding and subtracting the compound angle formulas gives a second family of identities. These turn a sum or difference of two sines or cosines into a product, and a product back into a sum. They are useful for exact values and for simplifying.
Rules to remember
- sin 2θ = 2 sin θ cos θ and tan 2θ = 2 tan θ/(1 − tan2θ)
- cos 2θ = cos2θ − sin2θ = 2 cos2θ − 1 = 1 − 2 sin2θ
- sin2(θ/2) = (1 − cos θ)/2 and cos2(θ/2) = (1 + cos θ)/2
- sin 3θ = 3 sin θ − 4 sin3θ and cos 3θ = 4 cos3θ − 3 cos θ
- sin P + sin Q = 2 sin((P + Q)/2) cos((P − Q)/2) and sin P − sin Q = 2 cos((P + Q)/2) sin((P − Q)/2)
- cos P + cos Q = 2 cos((P + Q)/2) cos((P − Q)/2) and cos P − cos Q = −2 sin((P + Q)/2) sin((P − Q)/2)
- 2 sin α cos β = sin(α + β) + sin(α − β); 2 cos α cos β = cos(α + β) + cos(α − β); 2 sin α sin β = cos(α − β) − cos(α + β)
Common mistake
sin 2θ is not 2 sin θ, and cos 3θ is not 3 cos θ. Also remember the minus sign in cos P − cos Q = −2 sin((P + Q)/2) sin((P − Q)/2).
Worked example
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