Double-angle and half-angle identities are special instances of the sum and distinct formulas for sine and cosine. First, using the sine's sum individuality,
sin 2α = sin (α + α)
sin 2α = sin α cos α + cos α sin α
sin 2α = 2 sin α cos α
Likewise, consider the cosine.
cos 2α = cos (α + α)
cos 2α = cos α cos α - sin α sin α
cos 2α = cos2 α - sin2 α
In this answer, we will learn how to find the exact values of the trig functions for angles that are half or double of other angles in this concept. The Double-Angle (2a) Formulas will be introduced here.
Double-Angle (2a) Formulas:
cos 2α = cos2 α - sin2 α
= 2 cos2 α – 1
= 1 - sin2 α
sin 2α = 2 sin α cos α
tan 2α = 2 tan α/1- tan2 α
For cos 2α and tan α/2, any formula can be used to resolve for the exact value.
Let us find the exact value of cos π/8.
π/8 is half of π/4 and in the first quadrant.
cos (1/2 * π/4) = √1 + cos π/4 / 2
= √1 + √2/2 / 2
= √1/2 * 2 + √2/2
= √2 + √2 / 2
Now, let us find the exact value of sin 2α if cos α = - 4/5 and 3π/2 ≤ α < 2π.
In order to use the sine double-angle formula, we also need to find sin α, which would be 3/5 because α is in the 4th quadrant.
sin 2α = 2 sin α cos α
= 2 * 3/5 * -4/5
= -24/25
Finally, let us determine the exact value of tan 2α for α from the prior problem.
Use tan α = sin α/ cos α = 3/5 / -4/5 = -3/4 to solve for tan 2α.
tan 2α = 2 * -3/4 / 1 – (-3/4)2 = -3/2 / 7/16 = -3/2 * 16/7 = -24/7
In this answer, we will learn how to find the exact values of the trig functions for angles that are half or double of other angles in this concept. The Half-Angle (a/2) Formulas will be introduced here.
Half-Angle (a/2) Formulas:
sin α/2 = ± √1 – cos α / 2
cos α/2 = ± √1 + cos α / 2
tan α/2 = 1 – cos α / sin α
= sin α / 1 + cos α
The signs of sin α/2 and cos α/2 rely on which quadrant α/2 lies.
The sine half-angle formula is deduced as follows:
sin2 θ = 1 – cos (2θ) / 2
sin2 α/2 = 1 – (cos 2 * α/2) / 2
= 1 - cos α / 2
sin α/2 = ± √1 - cos α / 2
We have used the half-angle formula for cosine to generate the formula.
cos2 θ = 1 + cos (2θ) / 2
cos2 α/2 = 1 + (cos 2 * α/2) / 2
= 1 + cos α / 2
cos (π/2) = ± √1 + cos α / 2
For the tangent identity, we have
tan2 θ = 1 – cos (2θ) / 1 + cos (2θ)
tan2 (α/2) = 1 – cos (2 * α/2) / 1 + cos (2 * α/2)
tan (α/2) = ± √1 – cos α / 1 + cos α
The same steps used to derive the sum and difference formulas are used to establish identities using the double-angle formulas. Rewrite the more complex side of the equation until it matches the other side.
Using double-angle formulas, confirm the following identity:
1 + sin(2θ) = (sin θ + cos θ)2
Solution:
We will start with the right side of the equal sign and rewrite the expression until it matches the left.
(sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ
= (sin2 θ + cos2 θ) + 2 sin θ cos θ
= 1 + sin θ cos θ
= 1 + sin (2θ)
Analysis
This procedure is simple if we remember the perfect square formula from algebra:
(a ± b)2 = a2 ± 2ab ± b2
Where a = sin θ and b = cos θ. Recognizing patterns is an important part of being successful in mathematics. The algebra remains constant even if the terms or symbols change.
Double-Angle Identities:
Sum identity for sine:
sin (x + y) = (sin x)(cos y) + (cos x)(sin y)
sin (x + x) = (sin x)(cos x) + (cos x)(sin x) (replace y with x)
sin 2x = 2 sin x cos x
Example 1: Verify, (sin x + cos x)2 = 1 + sin 2x:
Answer
(sin x + cos x)2 = 1 + sin 2x
(sin x + cos x)(sin x + cos x) = 1 + sin 2x
sin2 x + sin x cos x + sin x cos x + cos2 x = 1 + sin 2x
sin2 x + 2sin x cos x + cos2 x = 1 + sin 2x (combine like terms)
sin2 x + sin 2x + cos2 x = 1 + sin 2x (substitution: double-angle identity)
sin2 x + cos2 x + sin 2x = 1 + sin 2x
1 + sin 2x = 1 + sin 2x (Pythagorean identity)
Thus, 1 + sin 2x = 1 + sin 2x, is verifiable.
Half-Angle Identities:
Example 2: Without using a calculator, find the exact value for tan 30 degrees and use the half-angle identities.
Answer
tan 30 degrees = tan 60 degrees/ 2
= sin 60/ (1 + cos 60)
= ( √3 / 2) / (1 + 1/ 2) +
= ( √3 / 2) / (3 / 2)
= ( √3 / 2) (2 / 3) ×
= √3/3
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