Double angle formulas in trigonometry
The double angle formulas are the special trigonometric formulas involving the angles with the number 2. They are highly useful while solving complex trigonometric entities. Instead of deriving them while solving the problems, its better to remember this formulas so as to apply them directly and avoiding lot of unnecessary steps. Lets have a look at some of the important trigonometric formulas with double angle.

sin2θ = 2sinθcosθ = 2tanθ / (1+tan^{2}θ).

cos2θ = cos^{2}θ – sin^{2}θ = (1tan^{2}θ) / (1+tan^{2}θ) = 2cos^{2}θ – 1 = 1  2sin^{2}θ.

tan2θ = 2tanθ / (1tan^{2}θ).
This double angles formula makes the task of solving hard trigonometric problems very easy. Lets study the double angle formula proof for some of them.
Study of double angle formula proof
Lets replace 2θ by (θ+θ) and try to prove them all.
(1) sin2θ = sin(θ+θ). But we know that sin(A + B) = sinAcosB + cosAsinB. Hence sin2θ = sinθcosθ + cosθsinθ = 2sinθcosθ. Thus first double angle formula is proved. So lets move to the second one.
(2) cos2θ = cos(θ+θ). We have cos(A + B) = cosAcosB – sinAsinB. Applying this formula, we get the solution very easily. cos2θ = cosθcosθ – sinθsinθ = cos^{2}θ – sin^{2}θ. Similarly other variants of the formula can be proved.
(3) tan2θ = tan(θ+θ) = (tanθ + tanθ) / (1 – tanθtanθ) = 2tanθ / (1 – tan^{2}θ).
Some important double angle formula examples
The double angle formula examples given below illustrates the use of this formulas in solving trigonometric problems.
(1) Solve for θ if cos2θ = sinθ, π ≦ θ < π:
From the formula cos2θ = 1 – 2sin^{2}θ, 1 – 2sin^{2}θ = sinθ or 2sin^{2}θ + sinθ – 1 = 0. This equation can further be factorized to yield (2sinθ – 1)(sinθ + 1) = 0. This gives two value of sinθ as ½ or 1. Hence θ could be either π/6 or π/2.
The double angle formulas examples given above explains how to apply this formulas for solving the trig equations with ease.
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