sin^2(x) + cos^2(x) = 1

• 1 + cot^2(x) = csc^2(x)
• tan^2(x) + 1 = sec^2(x)

sin(u + v) = sin u cos v + cos u sin v

cos(u + v) = cos u cos v - sin u sin v

• sin(u - v) = sin (u) cos(v) - cos(u) sin(v)
• cos(u - v) = cos u cos(v) + sin u sin(v)
• tan(u + v) = (tan u + tan v) / ( 1 - tan u tan v)

Flip-and-shift-pi/2-units identities

(or the cofunction identities)

sin(pi / 2 - x) = cos(x)        cos(pi / 2 - x) = sin(x)

tan(pi / 2 - x) = cot x          cot(pi / 2 - x) = tan x

sec(pi / 2 - x) = csc x        csc(pi / 2 - x) = sec x

Shifting left pi/2 gives you

sin(x+pi/2) = cos(x)        cos(x+pi/2) = -sin(x)

From sin(u+v) = sin u cos v + cos u sin v let u = x and v = x to get

• sin(2x) = 2 sin x cos x

From cos(u+v) = cos u cos v - sin u sin v let u = x and v = x to get

• cos(2x) = cos^2(x) - sin^2(x) = 1 - 2 sin^2(x) = 2 cos^2(x) - 1

(Last two steps: substitute 1 - sin^2(x) for cos^2(x) then in cos^2(x) - sin^2(x) , then substitute 1 - cos^2(x) for sin^2(x) in cos^2(x) - sin^2(x) )

From tan(u+v) = (tan u + tan v) / ( 1 - tan u tan v) substitute u = x and v = x to get

• tan(2x) = 2 tan x / (1 - tan^2(x)).

From cos(2 u) = 1 - 2 sin^2(u), let u = x/2 to get sin(x) = 1 - 2 sin^2(x/2), which you solve for sin(x/2) to get

• sin(x/2) = +-sqrt( (1 - cos x) / 2)

From cos(2 u) = 2 cos^2(u) - 1, let u = x/2 to get cos(x) = 2 cos^2(x/2) - 1, which you solve for cos(x/2) to get

• cos(x/2) = +-sqrt( (1 + cos x) / 2)

Since tan(x/2) = sin(x/2) / cos(x/2), use the preceding two identities to get

• tan(x/2) = +- sqrt( (1 - cos x) / (1 + cos x) ) = sin x / (1 + cos x) = (1 - cos x) / sin x