20. Calculus

In calculus the relations stated below require angles to be measured in radians; the relations would become more complicated if angles were measured in another unit such as degrees. If the trigonometric functions are defined in terms of geometry, along with the definitions of arc length and area, their derivatives can be found by verifying two limits. The first is:

${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1,}$

verified using the unit circle and squeeze theorem. The second limit is:

${\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos x}{x}}=0,}$

verified using the identity tan x/2 = 1 − cos x/sin x. Having established these two limits, one can use the limit definition of the derivative and the addition theorems to show that (sin x)′ = cos x and (cos x)′ = −sin x. If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term.

${\displaystyle {\mathrm {d} \over \mathrm {d} x}\sin x=\cos x}$

The rest of the trigonometric functions can be differentiated using the above identities and the rules of differentiation:

${\displaystyle {\begin{aligned}{\mathrm {d} \over \mathrm {d} x}\sin x&=\cos x,&{\mathrm {d} \over \mathrm {d} x}\arcsin x&={1 \over {\sqrt {1-x^{2}}}}\\\\{\mathrm {d} \over \mathrm {d} x}\cos x&=-\sin x,&{\mathrm {d} \over \mathrm {d} x}\arccos x&={-1 \over {\sqrt {1-x^{2}}}}\\\\{\mathrm {d} \over \mathrm {d} x}\tan x&=\sec ^{2}x,&{\mathrm {d} \over \mathrm {d} x}\arctan x&={1 \over 1+x^{2}}\\\\{\mathrm {d} \over \mathrm {d} x}\cot x&=-\csc ^{2}x,&{\mathrm {d} \over \mathrm {d} x}\operatorname {arccot} x&={-1 \over 1+x^{2}}\\\\{\mathrm {d} \over \mathrm {d} x}\sec x&=\tan x\sec x,&{\mathrm {d} \over \mathrm {d} x}\operatorname {arcsec} x&={1 \over |x|{\sqrt {x^{2}-1}}}\\\\{\mathrm {d} \over \mathrm {d} x}\csc x&=-\csc x\cot x,&{\mathrm {d} \over \mathrm {d} x}\operatorname {arccsc} x&={-1 \over |x|{\sqrt {x^{2}-1}}}\end{aligned}}}$

The integral identities can be found in List of integrals of trigonometric functions. Some generic forms are listed below.

${\displaystyle \int {\frac {\mathrm {d} u}{\sqrt {a^{2}-u^{2}}}}=\sin ^{-1}\left({\frac {u}{a}}\right)+C}$

${\displaystyle \int {\frac {\mathrm {d} u}{a^{2}+u^{2}}}={\frac {1}{a}}\tan ^{-1}\left({\frac {u}{a}}\right)+C}$

${\displaystyle \int {\frac {\mathrm {d} u}{u{\sqrt {u^{2}-a^{2}}}}}={\frac {1}{a}}\sec ^{-1}\left|{\frac {u}{a}}\right|+C}$

Implications

The fact that the differentiation of trigonometric functions (sine and cosine) results in linear combinations of the same two functions is of fundamental importance to many fields of mathematics, including differential equations and Fourier transforms.

Some differential equations satisfied by the sine function

Let i = √−1 be the imaginary unit and let ∘ denote composition of differential operators. Then for every odd positive integer n,

${\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\binom {n}{k}}&\left({\frac {\mathrm {d} }{\mathrm {d} x}}-\sin x\right)\circ \left({\frac {\mathrm {d} }{\mathrm {d} x}}-\sin x+i\right)\circ \cdots \\&\qquad \cdots \circ \left({\frac {\mathrm {d} }{\mathrm {d} x}}-\sin x+(k-1)i\right)(\sin x)^{n-k}=0.\end{aligned}}}$

(When k = 0, then the number of differential operators being composed is 0, so the corresponding term in the sum above is just (sin x)ⁿ.) This identity was discovered as a by-product of research in medical imaging