Differentiation of trigonometric functions

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Short description: Mathematical process of finding the derivative of a trigonometric function
Function Derivative
sin(x) cos(x)
cos(x) sin(x)
tan(x) sec2(x)
cot(x) csc2(x)
sec(x) sec(x)tan(x)
csc(x) csc(x)cot(x)
arcsin(x) 11x2
arccos(x) 11x2
arctan(x) 1x2+1
\arccot(x) 1x2+1
\arcsec(x) 1|x|x21
\arccsc(x) 1|x|x21

The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. For example, the derivative of the sine function is written sin′(a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle.

All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x). Knowing these derivatives, the derivatives of the inverse trigonometric functions are found using implicit differentiation.

Proofs of derivatives of trigonometric functions

Limit of sin(θ)/θ as θ tends to 0

Circle, centre O, radius 1

The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < ½ π in the first quadrant.

In the diagram, let R1 be the triangle OAB, R2 the circular sector OAB, and R3 the triangle OAC.

The area of triangle OAB is:

Area(R1)=12 |OA| |OB|sinθ=12sinθ.

The area of the circular sector OAB is:

Area(R2)=12θ.

The area of the triangle OAC is given by:

Area(R3)=12 |OA| |AC|=12tanθ.

Since each region is contained in the next, one has:

Area(R1)<Area(R2)<Area(R3)12sinθ<12θ<12tanθ.

Moreover, since sin θ > 0 in the first quadrant, we may divide through by ½ sin θ, giving:

1<θsinθ<1cosθ1>sinθθ>cosθ.

In the last step we took the reciprocals of the three positive terms, reversing the inequities.

Squeeze: The curves y = 1 and y = cos θ shown in red, the curve y = sin(θ)/θ shown in blue.

We conclude that for 0 < θ < ½ π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:

limθ0+sinθθ=1.

For the case where θ is a small negative number –½ π < θ < 0, we use the fact that sine is an odd function:

limθ0sinθθ = limθ0+sin(θ)θ = limθ0+sinθθ = limθ0+sinθθ = 1.

Limit of (cos(θ)-1)/θ as θ tends to 0

The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of θ is unimportant.

limθ0cosθ1θ = limθ0(cosθ1θ)(cosθ+1cosθ+1) = limθ0cos2θ1θ(cosθ+1).

Using cos2θ – 1 = –sin2θ, the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:

limθ0cosθ1θ = limθ0sin2θθ(cosθ+1) = (limθ0sinθθ)(limθ0sinθcosθ+1) = (1)(02)=0.

Limit of tan(θ)/θ as θ tends to 0

Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:

limθ0tanθθ = (limθ0sinθθ)(limθ01cosθ) = (1)(1) = 1.

Derivative of the sine function

We calculate the derivative of the sine function from the limit definition:

ddθsinθ=limδ0sin(θ+δ)sinθδ.

Using the angle addition formula sin(α+β) = sin α cos β + sin β cos α, we have:

ddθsinθ=limδ0sinθcosδ+sinδcosθsinθδ=limδ0(sinδδcosθ+cosδ1δsinθ).

Using the limits for the sine and cosine functions:

ddθsinθ=(1)cosθ+(0)sinθ=cosθ.

Derivative of the cosine function

From the definition of derivative

We again calculate the derivative of the cosine function from the limit definition:

ddθcosθ=limδ0cos(θ+δ)cosθδ.

Using the angle addition formula cos(α+β) = cos α cos β – sin α sin β, we have:

ddθcosθ=limδ0cosθcosδsinθsinδcosθδ=limδ0(cosδ1δcosθsinδδsinθ).

Using the limits for the sine and cosine functions:

ddθcosθ=(0)cosθ(1)sinθ=sinθ.

From the chain rule

To compute the derivative of the cosine function from the chain rule, first observe the following three facts:

cosθ=sin(π2θ)
sinθ=cos(π2θ)
ddθsinθ=cosθ

The first and the second are trigonometric identities, and the third is proven above. Using these three facts, we can write the following,

ddθcosθ=ddθsin(π2θ)

We can differentiate this using the chain rule. Letting f(x)=sinx,  g(θ)=π2θ, we have:

ddθf(g(θ))=f(g(θ))g(θ)=cos(π2θ)(01)=sinθ.

Therefore, we have proven that

ddθcosθ=sinθ.

Derivative of the tangent function

From the definition of derivative

To calculate the derivative of the tangent function tan θ, we use first principles. By definition:

ddθtanθ=limδ0(tan(θ+δ)tanθδ).

Using the well-known angle formula tan(α+β) = (tan α + tan β) / (1 - tan α tan β), we have:

ddθtanθ=limδ0[tanθ+tanδ1tanθtanδtanθδ]=limδ0[tanθ+tanδtanθ+tan2θtanδδ(1tanθtanδ)].

Using the fact that the limit of a product is the product of the limits:

ddθtanθ=limδ0tanδδ×limδ0(1+tan2θ1tanθtanδ).

Using the limit for the tangent function, and the fact that tan δ tends to 0 as δ tends to 0:

ddθtanθ=1×1+tan2θ10=1+tan2θ.

We see immediately that:

ddθtanθ=1+sin2θcos2θ=cos2θ+sin2θcos2θ=1cos2θ=sec2θ.

From the quotient rule

One can also compute the derivative of the tangent function using the quotient rule.

ddθtanθ=ddθsinθcosθ=(sinθ)cosθsinθ(cosθ)cos2θ=cos2θ+sin2θcos2θ

The numerator can be simplified to 1 by the Pythagorean identity, giving us,

1cos2θ=sec2θ

Therefore,

ddθtanθ=sec2θ

Proofs of derivatives of inverse trigonometric functions

The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of. Using implicit differentiation and then solving for dy/dx, the derivative of the inverse function is found in terms of y. To convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting θ be y. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.

Differentiating the inverse sine function

We let

y=arcsinx

Where

π2yπ2

Then

siny=x

Taking the derivative with respect to x on both sides and solving for dy/dx:

ddxsiny=ddxx
cosydydx=1

Substituting cosy=1sin2y in from above,

1sin2ydydx=1

Substituting x=siny in from above,

1x2dydx=1
dydx=11x2

Differentiating the inverse cosine function

We let

y=arccosx

Where

0yπ

Then

cosy=x

Taking the derivative with respect to x on both sides and solving for dy/dx:

ddxcosy=ddxx
sinydydx=1

Substituting siny=1cos2y in from above, we get

1cos2ydydx=1

Substituting x=cosy in from above, we get

1x2dydx=1
dydx=11x2

Alternatively, once the derivative of arcsinx is established, the derivative of arccosx follows immediately by differentiating the identity arcsinx+arccosx=π/2 so that (arccosx)=(arcsinx).

Differentiating the inverse tangent function

We let

y=arctanx

Where

π2<y<π2

Then

tany=x

Taking the derivative with respect to x on both sides and solving for dy/dx:

ddxtany=ddxx

Left side:

ddxtany=sec2ydydx=(1+tan2y)dydx using the Pythagorean identity

Right side:

ddxx=1

Therefore,

(1+tan2y)dydx=1

Substituting x=tany in from above, we get

(1+x2)dydx=1
dydx=11+x2

Differentiating the inverse cotangent function

We let

y=\arccotx

where 0<y<π. Then

coty=x

Taking the derivative with respect to x on both sides and solving for dy/dx:

ddxcoty=ddxx

Left side:

ddxcoty=csc2ydydx=(1+cot2y)dydx using the Pythagorean identity

Right side:

ddxx=1

Therefore,

(1+cot2y)dydx=1

Substituting x=coty,

(1+x2)dydx=1
dydx=11+x2

Alternatively, as the derivative of arctanx is derived as shown above, then using the identity arctanx+\arccotx=π2 follows immediately thatddx\arccotx=ddx(π2arctanx)=11+x2

Differentiating the inverse secant function

Using implicit differentiation

Let

y=\arcsecx |x|1

Then

x=secy y[0,π2)(π2,π]
dxdy=secytany=|x|x21

(The absolute value in the expression is necessary as the product of secant and tangent in the interval of y is always nonnegative, while the radical x21 is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)

dydx=1|x|x21

Using the chain rule

Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the chain rule.

Let

y=\arcsecx=arccos(1x)

Where

|x|1 and y[0,π2)(π2,π]

Then, applying the chain rule to arccos(1x):

dydx=11(1x)2(1x2)=1x211x2=1x2x21x2=1x2x21=1|x|x21

Differentiating the inverse cosecant function

Using implicit differentiation

Let

y=\arccscx |x|1

Then

x=cscy  y[π2,0)(0,π2]
dxdy=cscycoty=|x|x21

(The absolute value in the expression is necessary as the product of cosecant and cotangent in the interval of y is always nonnegative, while the radical x21 is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)

dydx=1|x|x21

Using the chain rule

Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the chain rule.

Let

y=\arccscx=arcsin(1x)

Where

|x|1 and y[π2,0)(0,π2]

Then, applying the chain rule to arcsin(1x):

dydx=11(1x)2(1x2)=1x211x2=1x2x21x2=1x2x21=1|x|x21

See also

References

Bibliography