• Temporary trigonometry page

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Using the Cartesian coordinate system, the circle C is centred at the origin O and has radius 1. P is any point on that circle and has coordinates (a, b), and \theta is the angle between OP and the positive x-axis, measured anticlockwise. Then the following two basic trigonometrical functions are defined:

\sin \theta = b, \; \cos \theta = a

Furthermore, the following four functions are defined:

\displaystyle \tan \theta = \frac{\sin\theta}{\cos\theta}, \;\;\;\; \cot \theta = \frac{\cos\theta}{\sin\theta}, \;\;\;\; \sec \theta = \frac{1}{\cos\theta}, \;\;\;\; \text{cosec} \; \theta = \csc \theta = \frac{1}{\sin\theta}

Clearly tan and sec are not defined at any values of θ for which cos θ = 0; similarly, cot θ and cosec θ are only defined when sin θ does not equal 0.

Note that the notation "csc θ" will be preferred to "cosec θ" in this article.


Basic identities

Properties of the functions

By considering θ and (-θ) in the definitions of sin and cos, it is easy to see that

\sin (-\theta ) = -\sin\theta , \;\;\;\; \cos (-\theta ) = \cos \theta ,

that is, sin is an odd function, and cos is an even function. This makes tan, cot and csc odd functions, and sec an even function, i.e.

\tan (-\theta ) = -\tan\theta , \;\; \cot (-\theta ) = -\cot \theta , \;\; \csc (-\theta ) = \csc\theta , \;\; \sec (-\theta ) = \sec \theta .

\sin (30) = cos (60) = {1\over2}

\sin (60) = cos (30) = {\sqrt3\over2}

\sin (45) = cos (45) = {\sqrt2\over2}

\tan (30) = {\sqrt3\over3}

\tan (60) = {\sqrt3}

\tan (45) = 1

Basic relationships between the functions

It is obvious from our definitions of sin and cos, and by reflective symmetry, that

\sin \theta = \cos (90 - \theta) , \;\;\;\; \cos \theta = \sin (90 - \theta).

By dividing the first identity by the second, or the second by the first, respectively, we get

\tan x = \cot ({\pi\over 2} - x) , \;\;\;\; \cob x = \tan ({\pi\over 2} - x).

It is also immediately obvious, by taking the reciprocal of each of the first two identities, that

\csc x = \sec ({\pi\over 2} - x) , \;\;\;\; \sec x = \csc ({\pi\over 2} - x).

Circular identities

From our definitions of sin and cos, and using Pythagoras' theorem, it is easily seen that

\sin^2x + \cos^2x = 1.

By dividing through by sin²x and cos²x respectively, we end up with the two identities given below. Although sin²x and cos²x can equal 0 (exactly when sin x = 0 and cos x = 0, respectively), our definitions of tan, sec, cosec and cot allow us to divide through.

1 + \cot^2x = \csc^2 x

\tan^2x + 1 = \sec^2x

Addition identities

[insert proof of sin(A+B) formula here :p]

\sin (A+B) = \sin A\cos B + \sin B\cos A

By considering sin(A - B) = sin(A + (-B)):

\sin (A-B) = \sin A\cos (-B) + \sin (-B)\cos A

\sin (A-B) = \sin A\cos B - \sin B\cos A

By considering cos(A + B) = sin(π/2 - (A + B)) = sin((π/2 - A) - B):

\cos(A+B) = \sin({\pi\over 2} - A) \cos B - \sin B \cos ({\pi\over 2} - A)

\cos(A+B) = \cos A\cos B - \sin A\sin B

By considering cos(A - B) = cos(A + (-B)):

\cos(A-B) = \cos A\cos (-B) - \sin A\sin (-B)

\cos(A-B) = \cos A\cos B + \sin A\sin B

It is immediately obvious that

\displaystyle \tan(A+B) = \frac{\sin(A+B)}{\cos(A+B)} = \frac{\sin A\cos B + \sin B\cos A}{\cos A\cos B - \sin A\sin B} ,

which simplifies to the following identity when the numerator and denominator are divided by cos A and cos B:

\displaystyle \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} .


\displaystyle \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B} .

We can derive the addition identities for cot in a similar way.

\displaystyle \cot(A+B) = \frac{\cot A + \cot B}{\cot A\cot B - 1}

\displaystyle \cot(A-B) = \frac{\cot A - \cot B}{\cot A\cot B + 1}

The addition identities for sec and csc are derived by taking the reciprocal of the identities for cos and sin, respectively, and multiplying the numerator and denominator by (sec A sec B csc A csc B).

\displaystyle \sec(A+B) = \frac{\sec A\sec B\csc A\csc B}{\csc A\csc B - \sec A\sec B}

\displaystyle \sec(A-B) = \frac{\sec A\sec B\csc A\csc B}{\csc A\csc B + \sec A\sec B}

\displaystyle \csc(A+B) = \frac{\sec A\sec B\csc A\csc B}{\sec A\csc B + \sec B\csc A}

\displaystyle \csc(A-B) = \frac{\sec A\sec B\csc A\csc B}{\sec A\csc B - \sec B\csc A}

Double and half angle formulae

The double angle formulae are derived simply by setting B = A, and are as follows:

\sin 2A = 2\sin A\cos A

\cos 2A = \cos^2 A - \sin^2 A

\cos 2A = 2\cos^2 A - 1

\cos 2A = 1 - 2\sin^2 A

\displaystyle \tan 2A = \frac{2\tan A}{1 - \tan^2 A}

\displaystyle \cot 2A = \frac{2\cot A}{\cot^2 A - 1}

\displaystyle \sec 2A = \frac{\sec^2 A\csc^2 A}{\csc^2 A - \sec^2 A}

\csc 2A = {1\over 2}\sec A\csc A

The half angle formulae are equivalent to the double angle formulae, and are formed by replacing A by A/2 in the above formulae.


The half angle formula for sin can be written as follows, by multiplying the numerator and denominator by sec²θ.

\sin A = 2\sin {A\over 2}\cos {A\over 2} = \dfrac{2\tan {A\over 2}}{\sec^2 {A\over 2}} = \dfrac{2\tan {A\over 2}}{1 + \tan^2 {A\over 2}}

The same can be done with the half angle formula for cos.

\cos A = \cos^2 {A\over 2} - \sin^2 {A\over 2} = \dfrac{1 - \tan^2 {A\over 2}}{\sec^2 {A\over 2}} = \dfrac{1 - \tan^2 {A\over 2}}{1 + \tan^2 {A\over 2}}

Then, by writing t = tan(x/2), we arrive at the following six identities:

\sin x = \frac{2t}{1+t^2} , \;\; \cos x = \frac{1-t^2}{1+t^2} , \;\; \tan x = \frac{2t}{1-t^2} ,

\csc x = \frac{1+t^2}{2t} , \;\; \sec x = \frac{1+t^2}{1-t^2} , \;\; \cot x = \frac{1-t^2}{2t} .

The use of the substitution t = tan(x/2) is called t-substitution, and can be very useful in solving otherwise difficult or impossible integrals involving the simple trig functions.

Factor formulae

Product formulae

We know that

\sin (A+B) = \sin A\cos B + \sin B\cos A


\sin (A-B) = \sin A\cos B - \sin B\cos A .

Adding these two results in the product formula

\sin (A+B) + \sin (A-B) = 2\sin A\cos B.

Subtracting gives the formula

\sin (A+B) - \sin (A-B) = 2\sin B\cos A.

Similar manipulation of the addition identities for cos gives the two formulae

\cos (A+B) + \cos (A-B) = 2\cos A\cos B,

\cos (A+B) - \cos (A-B) = -2\sin A\sin B.

Sum formulae

The sum formulae are a direct result of the product formulae, by substituting A = P + Q, B = P - Q in the formulae

\sin (P-Q) = \sin P\cos Q - \sin Q\cos P ,   etc.

This gives:

\sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} ,

\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2} ,

\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2} ,

\cos A - \cos B = -2\sin \frac{A+B}{2} \sin \frac{A-B}{2} .



[insert proof that x --> 0 => sinx / x --> 1]

To differentiate sin x from first principles, let us consider the gradient m of the line between the points (x, sin x) and (x+o, sin(x+o)), where o is small.

\displaystyle m = \frac{\sin (x+o) - \sin x}{x+o - x} = \frac{\sin (x+o) - \sin x}{o}

\displaystyle = \frac{2\cos \tfrac{2x+o}{2} \sin \tfrac{o}{2}}{o} = \frac{\cos \tfrac{2x+o}{2} \sin \tfrac{o}{2}}{\tfrac{o}{2}}

\displaystyle =\cos (x + \tfrac{o}{2} ) \frac{\sin (o/2)}{o/2}

As o tends to 0, [sin(o/2)]/(o/2) tends to 1, and cos(x + o/2) tends to cos x, and so m tends to cos x; that is,

\dfrac{\text{d}}{\text{d} x} \sin x = \cos x.

Remember that cos x = sin(π/2 - x). Replacing x with (π/2 - x), and using the chain rule, in the above:

\dfrac{\text{d}}{\text{d} x} \sin ({\pi\over 2} - x) = -\cos ({\pi\over 2} - x).

That is,

\dfrac{\text{d}}{\text{d} x} \cos x = -\sin x.

Using the quotient rule:

\displaystyle \frac{\text{d}}{\text{d} x} \tan x = \frac{\text{d}}{\text{d} x} \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x.

\displaystyle \frac{\text{d}}{\text{d} x} \sec x = \frac{\text{d}}{\text{d} x} \frac{1}{\cos x} = \frac{\sin x}{\cos^2 x} = \sec x\tan x.

Deriving the final two by noting that cot x = tan(π/2 - x) and csc x = sec(π/2 - x) respectively:

\displaystyle \frac{\text{d}}{\text{d} x} \cot x = -\csc^2 x, \;\;\;\; \frac{\text{d}}{\text{d} x} \csc x = -\csc x\cot x.


Arbitrary constants are not added here.

As a direct result of the differentiation results,

\displaystyle \int \sin x \text{ d} x = -\cos x,

\displaystyle \int \cos x \text{ d} x = \sin x.


\displaystyle \int \tan x \text{ d} x = \int \frac{\sin x}{\cos x} \text{ d} x = -\ln \cos x = \ln \sec x,

\displaystyle \int \cot x \text{ d} x = \int \frac{\cos x}{\sin x} \text{ d} x = \ln \sin x.

The other two functions can be integrated with a small amount of manipulation.

\displaystyle \int \sec x \text{ d} x = \int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \text{d} x

\displaystyle = \int \frac{\sec^2 x+\sec x\tan x}{\sec x+\tan x} \text{d} x = \ln (\sec x+\tan x)

as the numerator is the derivative of the denominator.

\displaystyle \int \csc x \text{ d} x = \int \frac{\csc x(\csc x+\cot x)}{\csc x+\cot x} \text{ d} x

\displaystyle =- \int \frac{-\csc^2 x-\csc x\cot x}{\csc x+\cot x} \text{ d} x = -\ln (\csc x+\cot x)

as the numerator is the derivative of the denominator.

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