Double-angle formulas are formulas in trigonometry to solve trigonometric functions where the angle is a multiple of 2, i.e., in the form of (2θ).
Double Angle Formulas Derivation
Trigonometric formulae known as the "double angle identities" define the trigonometric functions of twice an angle in terms of the trigonometric functions of the angle itself. I'll be obtaining the sine, cosine, and tangent double angle identities here.
Derivation of Sine Double Angle Formula
Sine Double Angle Identity:
sin(2θ) = 2 sinθ cosθ
Start with the sum-to-product identity for sine:
sin (A + B) =sin A cos B + cos A sin B
Let A = θ and B = θ
sin(2θ) = sin(θ+θ)
sin(2θ) = sinθ cosθ + cosθ sinθ
sin(2θ) = 2sinθ cosθ
Derivation of Cos Double Angle Formula
Cosine Double Angle Identity:
cos(2θ) = cos2(θ) - sin2(θ)
Start with the sum-to-product identity for cosine:
cos (A + B) = cos A cos B - sin A sin B
Let A = θ and B = θ
cos(2θ) = cos(θ+θ)
cos(2θ) = cosθcosθ - sinθ sinθ
cos(2θ) = cos2θ - sin2θ
Derivation of Tan Double Angle Formula
Tangent Double Angle Identity:
tan(2θ) = 2tanθ / [1 - tan2θ]
Use the quotient identity for tangent:
tan(A+B) = [tan A + tan B] / [1 - tan A tan B]
Let A=θ and B=θ
tan(2θ) = tan(θ+θ)
tan(2θ) = [tan(θ) + tan(θ)] / [1 - tan(θ)tan(θ)]
tan(2θ) = 2tan(θ) / [1 - tan2(θ)]
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Examples Using Double Angle Formulas
Example 1: Solve sin(2θ) = cos(θ) for θ
Solution:
sin(2θ) = cos(θ)
Using double angle identity for sine sin(2θ)=2sin(θ)cos(θ)), substitute:
2sin(θ)cos(θ) = cos(θ)
Now, divide both sides by cos(θ) (assuming cos(θ) ≠ 0
2sin(θ) = 1
Finally, solve for θ
sin(θ) = 1 /2
This implies θ = 30° or θ = 150°.
Example 2: Express tan(2x) in terms of tan(x):
Solution:
Using double angle identity for tangent
tan(2x) = 2tan(x) / {1 - tan2(x)}
This expression provides the tangent of twice the angle x in terms of the tangent of x.
Example 3: Use double angle identities to find the exact value of sin(120°)
Solution:
sin(2θ) = sin (240°)
Using, sin (180°+ θ) = - sin(θ)
We can rewrite expression,
-sin (60°) = - √3/2
Example 4: Prove the double angle identity for sine: sin(2θ) = 2sinθcosθ.
Solution:
Starting with (LHS)
sin(2θ) = sin(θ+θ)
sin(2θ) = sinθ cosθ + cosθ sinθ
Using trigonometric identity:
sin (a + b) = sin(a)cos(b) + cos(a)sin(b)
we get:
sin (θ + θ) = sin(θ)cos(θ) + cos(θ)sin(θ) = 2sin(θ)cos(θ)
Thus, LHS is equal to the right-hand side (RHS), and double angle identity for the sine is proved.
Practice Problems on Double Angle Formulas
Q1. Solve for sin(2θ) if sinθ = 3/5.
Q2. Express cos(2α) in terms of cos(α) if cos(α) = -4/7.
Q3. If tan(β) = 125, find the value of tan(2β).
Q4. Given that sin(ϕ) = 1/2 and ϕ is acute, determine cos(2ϕ).
Q5. Evaluate cot(2θ) if cotθ = -3/4.