Double Angle Trig Identity Solver

Double Angle Trig Identities calculator, computes sin (2u), cos (2u) and tan (2u) for given angle.

 Please enter the angle in degree:

 Result

The formula used

sin(2u) = 2 sinu cosu

cos(2u) = 1 - 2sin2u = 2cos2u - 1

tan(2u) = 2tanu / [ 1 - tan2u ]

Double angle formulas are allowing the expression of trigonometric functions of angles equal to 2u in terms of u, the double angle formulas can simplify the functions and gives ease to perform more complex calculations. The double angle formulas are useful for finding the values of unknown trigonometric functions. The double angle formulas can be elaborated to multiple angle functions using the angle sum formulas and then re-applying the double angle formulas.

Proof of Double Angle Identities

The Double Angle Formulas can be derived from Sum 2 angles (A & B) as follows:
$$\sin (A + B) = \sin A \, \cos B + \cos A \, \sin B ........(1)$$ $$\cos (A + B) = \cos A \, \cos B - \sin A \, \sin B ........(2)$$ $$\tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \, \tan B}..............(3)$$ lets assume angle A = B = u, apply this in equation (1) to derive the value for sin(2u) : $$\sin (u + u) = \sin u \, \cos u + \cos u \, \sin u$$ $$\sin 2u = 2\sin u \, \cos u$$ Simalarly apply angle A = B = u in equation (2) as follows to derive the value for cos(2u):
$$\cos (u + u) = \cos u \, \cos u - \sin u \, \sin u$$ $$\cos 2u = \cos^2 u - \sin^2 u...................(4)$$
We can redefine the Pythagorean Identity sin2u + cos2u = 1 as:
sin2u = 1 - cos2u
now apply this in equation (4)
$$\cos 2u = \cos^2 u - (1 - \cos^2 u)$$ $$\cos 2u = 2\cos^2 u - 1$$ We can also redefine the Pythagorean Identity sin2u + cos2u = 1 as:
cos2u = 1 - sin2u and now apply this in equation (4)
$$\cos 2u = (1 - \sin^2) - \sin^2 u$$ $$\cos 2u = 1 - 2\sin^2 u$$ lets conclude all the derived values for cos(2u) $$\cos 2u = \cos^2 u - \sin^2 u$$ $$\cos 2u = 2\cos^2 u - 1$$ $$\cos 2u = 1 - 2\sin^2 u$$ finally apply angle A = B = u in equation (3) as follows to derive the value for tan(2u):
$$\tan (u + u) = \dfrac{\tan u + \tan u}{1 - \tan u \, \tan u}$$ $$\tan 2u = \dfrac{2\tan u}{1 - \tan^2 u}$$