Explore: Cos Sin

sin(x)\cos(iy)+\cos(x)\sin(iy)\\&=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\\cos(x+iy)&=\cos(x)\cos(iy)-\sin(x)\sin(iy)\\&=\cos(x)\cosh(y)-i\sin

cos ⁡ x + i sin ⁡ x , {\displaystyle e^{ix}=\cos x+i\sin x,} where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are

cos ⁡ ( x − y ) = cos ⁡ x cos ⁡ y + sin ⁡ x sin ⁡ y {\displaystyle \cos(x-y)=\cos x\cos y+\sin x\sin y\,} and the added condition 0 < x cos ⁡ x < sin

hold: cos ⁡ a = cos ⁡ b cos ⁡ c + sin ⁡ b sin ⁡ c cos ⁡ A cos ⁡ A = − cos ⁡ B cos ⁡ C + sin ⁡ B sin ⁡ C cos ⁡ a cos ⁡ a = cos ⁡ A + cos ⁡ B cos ⁡ C sin ⁡

the matrix R = [ cos ⁡ θ − sin ⁡ θ sin ⁡ θ cos ⁡ θ ] {\displaystyle R={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}}

the case that ( cos ⁡ x + i sin ⁡ x ) n = cos ⁡ n x + i sin ⁡ n x , {\displaystyle {\big (}\cos x+i\sin x{\big )}^{n}=\cos nx+i\sin nx,} where i is the

formulae). sin ⁡ ( α + β ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β sin ⁡ ( α − β ) = sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β cos ⁡ ( α + β ) = cos ⁡ α cos ⁡ β − sin ⁡ α

equation cos ⁡ ω ∘ = sin ⁡ a − sin ⁡ ϕ × sin ⁡ δ cos ⁡ ϕ × cos ⁡ δ {\displaystyle \cos \omega _{\circ }={\dfrac {\sin a-\sin \phi \times \sin \delta }{\cos \phi

is omitted entirely. One can use cos − sin {\displaystyle \cos -\sin } instead of cos + sin {\displaystyle \cos +\sin } as the kernel. This transform differs

is sin 2 ⁡ θ + cos 2 ⁡ θ = 1. {\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1.} As usual, sin 2 ⁡ θ {\displaystyle \sin ^{2}\theta } means ( sin ⁡ θ