Limit of sin(θ)/θ as θ tends to 0 edit

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:

${\displaystyle \mathrm {Area} (R_{1})={\tfrac {1}{2}}\ |OA|\ |OB|\sin \theta ={\tfrac {1}{2}}\sin \theta \,.}$ 

The area of the circular sector OAB is:

${\displaystyle \mathrm {Area} (R_{2})={\tfrac {1}{2}}\theta \,.}$ 

The area of the triangle OAC is given by:

${\displaystyle \mathrm {Area} (R_{3})={\tfrac {1}{2}}\ |OA|\ |AC|={\tfrac {1}{2}}\tan \theta \,.}$ 

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

${\displaystyle {\text{Area}}(R_{1})<{\text{Area}}(R_{2})<{\text{Area}}(R_{3})\implies {\tfrac {1}{2}}\sin \theta <{\tfrac {1}{2}}\theta <{\tfrac {1}{2}}\tan \theta \,.}$ 

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

${\displaystyle 1<{\frac {\theta }{\sin \theta }}<{\frac {1}{\cos \theta }}\implies 1>{\frac {\sin \theta }{\theta }}>\cos \theta \,.}$ 

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

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:

${\displaystyle \lim _{\theta \to 0^{+}}{\frac {\sin \theta }{\theta }}=1\,.}$ 

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

${\displaystyle \lim _{\theta \to 0^{-}}\!{\frac {\sin \theta }{\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {\sin(-\theta )}{-\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {-\sin \theta }{-\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {\sin \theta }{\theta }}\ =\ 1\,.}$ 

Limit of (cos(θ)-1)/θ as θ tends to 0 edit

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.

${\displaystyle \lim _{\theta \to 0}\,{\frac {\cos \theta -1}{\theta }}\ =\ \lim _{\theta \to 0}\left({\frac {\cos \theta -1}{\theta }}\right)\!\!\left({\frac {\cos \theta +1}{\cos \theta +1}}\right)\ =\ \lim _{\theta \to 0}\,{\frac {\cos ^{2}\!\theta -1}{\theta \,(\cos \theta +1)}}.}$ 

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:

${\displaystyle \lim _{\theta \to 0}\,{\frac {\cos \theta -1}{\theta }}\ =\ \lim _{\theta \to 0}\,{\frac {-\sin ^{2}\theta }{\theta (\cos \theta +1)}}\ =\ \left(-\lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}\right)\!\left(\lim _{\theta \to 0}\,{\frac {\sin \theta }{\cos \theta +1}}\right)\ =\ (-1)\left({\frac {0}{2}}\right)=0\,.}$ 

Limit of tan(θ)/θ as θ tends to 0 edit

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:

${\displaystyle \lim _{\theta \to 0}{\frac {\tan \theta }{\theta }}\ =\ \left(\lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}\right)\!\left(\lim _{\theta \to 0}{\frac {1}{\cos \theta }}\right)\ =\ (1)(1)\ =\ 1\,.}$ 

Derivative of the sine function edit

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =\lim _{\delta \to 0}{\frac {\sin(\theta +\delta )-\sin \theta }{\delta }}.}$ 

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =\lim _{\delta \to 0}{\frac {\sin \theta \cos \delta +\sin \delta \cos \theta -\sin \theta }{\delta }}=\lim _{\delta \to 0}\left({\frac {\sin \delta }{\delta }}\cos \theta +{\frac {\cos \delta -1}{\delta }}\sin \theta \right).}$ 

Using the limits for the sine and cosine functions:

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =(1)\cos \theta +(0)\sin \theta =\cos \theta \,.}$ 

Derivative of the cosine function edit

From the definition of derivative edit

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\cos \theta =\lim _{\delta \to 0}{\frac {\cos(\theta +\delta )-\cos \theta }{\delta }}.}$ 

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\cos \theta =\lim _{\delta \to 0}{\frac {\cos \theta \cos \delta -\sin \theta \sin \delta -\cos \theta }{\delta }}=\lim _{\delta \to 0}\left({\frac {\cos \delta -1}{\delta }}\cos \theta \,-\,{\frac {\sin \delta }{\delta }}\sin \theta \right).}$ 

Using the limits for the sine and cosine functions:

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\cos \theta =(0)\cos \theta -(1)\sin \theta =-\sin \theta \,.}$ 

From the chain rule edit

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

${\displaystyle \cos \theta =\sin \left({\tfrac {\pi }{2}}-\theta \right)}$ 

${\displaystyle \sin \theta =\cos \left({\tfrac {\pi }{2}}-\theta \right)}$ 

${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\sin \theta =\cos \theta }$ 

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

${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\cos \theta ={\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\sin \left({\tfrac {\pi }{2}}-\theta \right)}$ 

We can differentiate this using the chain rule. Letting ${\displaystyle f(x)=\sin x,\ \ g(\theta )={\tfrac {\pi }{2}}-\theta }$ , we have:

${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}f\!\left(g\!\left(\theta \right)\right)=f^{\prime }\!\left(g\!\left(\theta \right)\right)\cdot g^{\prime }\!\left(\theta \right)=\cos \left({\tfrac {\pi }{2}}-\theta \right)\cdot (0-1)=-\sin \theta }$ .

Therefore, we have proven that

${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\cos \theta =-\sin \theta }$ .

Derivative of the tangent function edit

From the definition of derivative edit

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =\lim _{\delta \to 0}\left({\frac {\tan(\theta +\delta )-\tan \theta }{\delta }}\right).}$ 

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =\lim _{\delta \to 0}\left[{\frac {{\frac {\tan \theta +\tan \delta }{1-\tan \theta \tan \delta }}-\tan \theta }{\delta }}\right]=\lim _{\delta \to 0}\left[{\frac {\tan \theta +\tan \delta -\tan \theta +\tan ^{2}\theta \tan \delta }{\delta \left(1-\tan \theta \tan \delta \right)}}\right].}$ 

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =\lim _{\delta \to 0}{\frac {\tan \delta }{\delta }}\times \lim _{\delta \to 0}\left({\frac {1+\tan ^{2}\theta }{1-\tan \theta \tan \delta }}\right).}$ 

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =1\times {\frac {1+\tan ^{2}\theta }{1-0}}=1+\tan ^{2}\theta .}$ 

We see immediately that:

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =1+{\frac {\sin ^{2}\theta }{\cos ^{2}\theta }}={\frac {\cos ^{2}\theta +\sin ^{2}\theta }{\cos ^{2}\theta }}={\frac {1}{\cos ^{2}\theta }}=\sec ^{2}\theta \,.}$ 

From the quotient rule edit

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

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\tan \theta ={\frac {\operatorname {d} }{\operatorname {d} \!\theta }}{\frac {\sin \theta }{\cos \theta }}={\frac {\left(\sin \theta \right)^{\prime }\cdot \cos \theta -\sin \theta \cdot \left(\cos \theta \right)^{\prime }}{\cos ^{2}\theta }}={\frac {\cos ^{2}\theta +\sin ^{2}\theta }{\cos ^{2}\theta }}}$ 

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

${\displaystyle {\frac {1}{\cos ^{2}\theta }}=\sec ^{2}\theta }$ 

Therefore,

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\tan \theta =\sec ^{2}\theta }$ 

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 edit

We let

${\displaystyle y=\arcsin x\,\!}$ 

Where

${\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}$ 

Then

${\displaystyle \sin y=x\,\!}$ 

Taking the derivative with respect to ${\displaystyle x}$  on both sides and solving for dy/dx:

${\displaystyle {d \over dx}\sin y={d \over dx}x}$ 

${\displaystyle \cos y\cdot {dy \over dx}=1\,\!}$ 

Substituting ${\displaystyle \cos y={\sqrt {1-\sin ^{2}y}}}$  in from above,

${\displaystyle {\sqrt {1-\sin ^{2}y}}\cdot {dy \over dx}=1}$ 

Substituting ${\displaystyle x=\sin y}$  in from above,

${\displaystyle {\sqrt {1-x^{2}}}\cdot {dy \over dx}=1}$ 

${\displaystyle {dy \over dx}={\frac {1}{\sqrt {1-x^{2}}}}}$ 

Differentiating the inverse cosine function edit

We let

${\displaystyle y=\arccos x\,\!}$ 

Where

${\displaystyle 0\leq y\leq \pi }$ 

Then

${\displaystyle \cos y=x\,\!}$ 

Taking the derivative with respect to ${\displaystyle x}$  on both sides and solving for dy/dx:

${\displaystyle {d \over dx}\cos y={d \over dx}x}$ 

${\displaystyle -\sin y\cdot {dy \over dx}=1}$ 

Substituting ${\displaystyle \sin y={\sqrt {1-\cos ^{2}y}}\,\!}$  in from above, we get

${\displaystyle -{\sqrt {1-\cos ^{2}y}}\cdot {dy \over dx}=1}$ 

Substituting ${\displaystyle x=\cos y\,\!}$  in from above, we get

${\displaystyle -{\sqrt {1-x^{2}}}\cdot {dy \over dx}=1}$ 

${\displaystyle {dy \over dx}=-{\frac {1}{\sqrt {1-x^{2}}}}}$ 

Alternatively, once the derivative of ${\displaystyle \arcsin x}$  is established, the derivative of ${\displaystyle \arccos x}$  follows immediately by differentiating the identity ${\displaystyle \arcsin x+\arccos x=\pi /2}$  so that ${\displaystyle (\arccos x)'=-(\arcsin x)'}$ .

Differentiating the inverse tangent function edit

We let

${\displaystyle y=\arctan x\,\!}$ 

Where

${\displaystyle -{\frac {\pi }{2}}<y<{\frac {\pi }{2}}}$ 

Then

${\displaystyle \tan y=x\,\!}$ 

Taking the derivative with respect to ${\displaystyle x}$  on both sides and solving for dy/dx:

${\displaystyle {d \over dx}\tan y={d \over dx}x}$ 

Left side:

${\displaystyle {d \over dx}\tan y=\sec ^{2}y\cdot {dy \over dx}=(1+\tan ^{2}y){dy \over dx}}$  using the Pythagorean identity

Right side:

${\displaystyle {d \over dx}x=1}$ 

Therefore,

${\displaystyle (1+\tan ^{2}y){dy \over dx}=1}$ 

Substituting ${\displaystyle x=\tan y\,\!}$  in from above, we get

${\displaystyle (1+x^{2}){dy \over dx}=1}$ 

${\displaystyle {dy \over dx}={\frac {1}{1+x^{2}}}}$ 

Differentiating the inverse cotangent function edit

We let

${\displaystyle y=\operatorname {arccot} x}$ 

where ${\displaystyle 0<y<\pi }$ . Then

${\displaystyle \cot y=x}$ 

Taking the derivative with respect to ${\displaystyle x}$  on both sides and solving for dy/dx:

${\displaystyle {\frac {d}{dx}}\cot y={\frac {d}{dx}}x}$ 

Left side:

${\displaystyle {d \over dx}\cot y=-\csc ^{2}y\cdot {dy \over dx}=-(1+\cot ^{2}y){dy \over dx}}$  using the Pythagorean identity

Right side:

${\displaystyle {d \over dx}x=1}$ 

Therefore,

${\displaystyle -(1+\cot ^{2}y){\frac {dy}{dx}}=1}$ 

Substituting ${\displaystyle x=\cot y}$ ,

${\displaystyle -(1+x^{2}){\frac {dy}{dx}}=1}$ 

${\displaystyle {\frac {dy}{dx}}=-{\frac {1}{1+x^{2}}}}$ 

Alternatively, as the derivative of ${\displaystyle \arctan x}$  is derived as shown above, then using the identity ${\displaystyle \arctan x+\operatorname {arccot} x={\dfrac {\pi }{2}}}$  follows immediately that

Differentiating the inverse secant function edit

Using implicit differentiation edit

Let

${\displaystyle y=\operatorname {arcsec} x\ \mid |x|\geq 1}$ 

Then

${\displaystyle x=\sec y\mid \ y\in \left[0,{\frac {\pi }{2}}\right)\cup \left({\frac {\pi }{2}},\pi \right]}$ 

${\displaystyle {\frac {dx}{dy}}=\sec y\tan y=|x|{\sqrt {x^{2}-1}}}$ 

(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 ${\displaystyle {\sqrt {x^{2}-1}}}$  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.)

${\displaystyle {\frac {dy}{dx}}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$ 

Using the chain rule edit

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

Let

${\displaystyle y=\operatorname {arcsec} x=\arccos \left({\frac {1}{x}}\right)}$ 

Where

${\displaystyle |x|\geq 1}$  and ${\displaystyle y\in \left[0,{\frac {\pi }{2}}\right)\cup \left({\frac {\pi }{2}},\pi \right]}$ 

Then, applying the chain rule to ${\displaystyle \arccos \left({\frac {1}{x}}\right)}$ :

${\displaystyle {\frac {dy}{dx}}=-{\frac {1}{\sqrt {1-({\frac {1}{x}})^{2}}}}\cdot \left(-{\frac {1}{x^{2}}}\right)={\frac {1}{x^{2}{\sqrt {1-{\frac {1}{x^{2}}}}}}}={\frac {1}{x^{2}{\frac {\sqrt {x^{2}-1}}{\sqrt {x^{2}}}}}}={\frac {1}{{\sqrt {x^{2}}}{\sqrt {x^{2}-1}}}}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$ 

Differentiating the inverse cosecant function edit

Using implicit differentiation edit

Let

${\displaystyle y=\operatorname {arccsc} x\ \mid |x|\geq 1}$ 

Then

${\displaystyle x=\csc y\ \mid \ y\in \left[-{\frac {\pi }{2}},0\right)\cup \left(0,{\frac {\pi }{2}}\right]}$ 

${\displaystyle {\frac {dx}{dy}}=-\csc y\cot y=-|x|{\sqrt {x^{2}-1}}}$ 

(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 ${\displaystyle {\sqrt {x^{2}-1}}}$  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.)

${\displaystyle {\frac {dy}{dx}}={\frac {-1}{|x|{\sqrt {x^{2}-1}}}}}$ 

Using the chain rule edit

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

Let

${\displaystyle y=\operatorname {arccsc} x=\arcsin \left({\frac {1}{x}}\right)}$ 

Where

${\displaystyle |x|\geq 1}$  and ${\displaystyle y\in \left[-{\frac {\pi }{2}},0\right)\cup \left(0,{\frac {\pi }{2}}\right]}$ 

Then, applying the chain rule to ${\displaystyle \arcsin \left({\frac {1}{x}}\right)}$ :

${\displaystyle {\frac {dy}{dx}}={\frac {1}{\sqrt {1-({\frac {1}{x}})^{2}}}}\cdot \left(-{\frac {1}{x^{2}}}\right)=-{\frac {1}{x^{2}{\sqrt {1-{\frac {1}{x^{2}}}}}}}=-{\frac {1}{x^{2}{\frac {\sqrt {x^{2}-1}}{\sqrt {x^{2}}}}}}=-{\frac {1}{{\sqrt {x^{2}}}{\sqrt {x^{2}-1}}}}=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$ 

• Calculus – Branch of mathematics
• Derivative – Instantaneous rate of change (mathematics)
• Differentiation rules – Rules for computing derivatives of functions
• General Leibniz rule – Generalization of the product rule in calculus
• Inverse functions and differentiation – Calculus identityPages displaying short descriptions of redirect targets
• Linearity of differentiation – Calculus property
• List of integrals of inverse trigonometric functions
• List of trigonometric identities – Equalities that involve trigonometric functions
• Trigonometry – Area of geometry, about angles and lengths

• Handbook of Mathematical Functions, Edited by Abramowitz and Stegun, National Bureau of Standards, Applied Mathematics Series, 55 (1964)