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# Trigonometrical Formulae

A collection of formulae covering, Addition and subtraction of Sin cos and tan; Product formulae ; the solution of equations and the half angle formulae and the Inverser Ratioi

If OP and OQ are unit radii which make a angles with the x axis of B and A respectively. Then the coordinates of P are (Cos B ; sin B) and for Q (cos A ; sin A). By inspection the angle POQ is of magnitude A - B.

Using the Pythagora theorem,

$PQ^2\;=\;(cos\,B\;-\;cos\,A)^2\;+\;(sin\,B\;-\;sin\,A)^2$

$=\;2\;-\;2\;cos\,B\;cos\,A\;-\;2\,sin\,B\;sin\,A$

Applying the cosine formula to triangle POQ:-

$PQ^2\;=\;1^2\;+\;1^2\;-\;2\times&space;1\times&space;1\times&space;cos\,(A\;-\;B)$

Equating equations (2) and (3)

$\mathbf{cos(A\;-\;B)\;=\;cos\,A\;cos\,B\;+\;sin\,A\;sin\,B}$

This equation applies for all values of A and B

Writing $\inline&space;(90^0\;+\;A)$ for A

$cos(90^0\;+\;A\;-\;B)\;=\;cos\,(90^0\;+\;A)\;cos\,B\;+\;sin\,(90^0\;+\;A)\;sin\,A$

$\therefore\;\;\;\;\;\;\mathbf{sin\,(A\;-\;B)\;=\;sin\,A\;cos\,B\;-\;cos\,A\;sin\,B}$

If B is replaced by - B and making use of the fact that cos B = cos(-B) and that - sin B = sin(-B).Then:-

$\mathbf{cos\,(A\;+\;B)\;=\;cos\,A\;cos\,B\;-\;sin\,A\;sin\,B}$

$\mathbf{sin\,(A\;+\;B)\;=\;sin\,A\;cos\,B\;+\;cos\,A\;sin\,B}$

Putting A = B

$\mathbf{cos\,2A\;=\;cos^2\,A\;-\;sin^2\;A}$

The above equation can be expressed in two different forms:-

$\mathbf{cos\,2A\;=\;2\;cos^2\,A\;-\;1}$

$\mathbf{cos\,2A\;=\;1\;-\;2\;sin^2\,A}$

Equation (8) can be treated the same way in which case:-

$\mathbf{sin\,2A\;=\;2\;sin\,A\;cos\,A}$

par Addition Formulae for the Tangent.

$\mathbf{tan\,(A\;-\;B)\;=\;\frac{tan\,A\;-\;tan\,B}{1\;+\;tan\,A\;tan\,B}}$
$tan\;(A\;+\;B)\;=\frac{sin\,(A\;+\;B)}{cos\;(A\;+\;B)}$

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

Divide the Numerator and the denominator by $\inline&space;\;cos\,A\;cos\,B$

$\therefore\;\;\;\;\;\;\mathbf{tan\,(A\;+\;B)\;=\;\frac{tan\,A\;+\;tan\,B}{1\;-\;tan\,A\;tan\,B}}$

If B is replaced in the above equation by - B

$\mathbf{tan\,(A\;-\;B)\;=\;\frac{tan\,A\;-\;tan\,B}{1\;+\;tan\,A\;tan\,B}}$

From equation (16) it can be seen that :-

$\mathbf{tan\,2A\;=\;\frac{2\;tan\,A}{1\;-\;tan^2\;A}}$

It is worth noting that :-

$tan\,(A\;+\;B\;+\;C)\;=\;\frac{tan\,A\;+\;tan\,(B\;+\;C)}{1\;-\;tan\,A\;tan\;(B\;+\;C)}$

$\therefore\;\;\;\;\;\;\;\mathbf{tan\,(A\;+\;B\;+\;C)\;=}$

$\mathbf{\frac{tan\,A\;+\;tan\,B\;+\;tan\,C\;-tan\,A\;tan\,B\;tan\,C}{1\;-\;tan\,A\;tan\,C\;-\;tan\,C\;tan\,A\;-\;tan\,A\;tan\,B}}$

This is a particular case of the more general formlua

$\mathbf{tan\,(A\;+\;B\;+\;C\;+\;....)\;=\;\frac{s_1\;-\;s_3\;+\;s_5\;-\;...}{1\;-\;s_2\;+\;s_4\;-\;...}}$

Where $\inline&space;&space;s_n$ stands for all the possible products of tan A ,tan B etc taken n at a time.

It follows from equation (21) that since the $\inline&space;&space;tan\,180^0\,=\;0$ and if A, B, C are the angles of a triangle then:-

$\mathbf{tan\,A\;+\;tan,B\;+\;tan\,C&space;\;=\;tan\,A\;tan\,B\;tan\,C}$

## Useful Formulae

$\mathbf{sin\,2A}\;=\;2\;sin\,A\;cos\,A\;=\;\frac{2\;tan\,A}{sec^2\,A}\;=\;\mathbf{\frac{2\;tan\,A}{1\;+\;tan^2\,A}}$

And

$\mathbf{cos\,2A}\;=\;2\;cos^2\,A\;-\;sin^2\,A\;=\;\frac{cos^2\,A\;-\;sin^2\,A}{cos^2\,A\;+\;sin^2\,A}\;=\;\mathbf{\frac{1\;-\;tan^2\,A}{1\;+\;tan^2\,A}}$

## The Product Formulae.

$since\;sin\,(A\;+\;B)\;=\;sin\,A\;cos\,B\;+\;cos\,A\;sin\,B$

$and\;\;\;sin\,(A\;-\;B)\;=\;sin\,A\;cos\,B\;-\;cos\,A\;sin\,B$

By adding the two above equations we get:-

$sin\,(A\;+\;B)\;+\;sin\,(A\;-\;B)\;=\;2\;sin\,A\;cos\,B$

And by subtraction:-

$sin\,(A\;+\;B)\;-\;sin\,(A\;-\;B)\;=\;2\;cos\,A\;sin\,B$

In these two new equations we can substitute (A + B) = X and (A - B) = Y from which :-

$A\;=\;\frac{1}{2}(X\;+\;Y)\;\;\;\;\;and\;\;\;\;\;B\;=\;\frac{1}{2}(X\;-\;Y)$

$\mathbf{sin\,X\;+\;sin\,Y\;=\;2\,sin\,\frac{1}{2}(X\;+\;Y)\times&space;cos\,\frac{1}{2}(X\;-\;Y)}$

And

$\mathbf{sin\,X\;-\;sin\,Y\;=\;2\,cos\,\frac{1}{2}(X\;+\;Y)\times&space;sin\,\frac{1}{2}(X\;-\;Y)}$

Proceeding in a similar way we get:-

$\mathbf{cos\,X\;+\;cos\,Y\;=\;2\,cos\,\frac{1}{2}(X\;+\;Y)\times&space;cos\,\frac{1}{2}(X\;-\;Y)}$

and
$\mathbf{cos\,X\;-\;cos\,Y\;=\;-\;2\,sin\,\frac{1}{2}(X\;+\;Y)\times&space;sin\,\frac{1}{2}(X\;-\;Y)}$

$\mathbf{=\;2\,sin\,\frac{1}{2}(X\;+\;Y)\times&space;sin\,\frac{1}{2}(Y\;-\;X)}$

## The Half Angle Formulae

By writing A = x/2 in formulae from the last sections :=

From equation (12)

$sin\,x\;=\;2\;sin\,\frac{1}{2}x\;cos\,\frac{1}{2}x$

And from (9) (10) (11)

$cos\,x\;=\;cos^2\,\frac{1}{2}x\;-\;sin^2\,\frac{1}{2}x\;=\;2\;cos^2\,\frac{1}{2}x\;-\;1\;=\;1\;-\;2\;sin^2\,\frac{1}{2}x$

and from equation (18)

$tan\,x\;=\;\frac{2\,tan\,\frac{1}{2}x}{1\;-\;tan^2\,\frac{1}{2}x}$

These formulae allow us to express the sine; cosine; and tangent of an angle in terms of the tangent of the half angle. It is therefore possible to write

$t\;=\;tan\,\frac{1}{2}x$

from which
$\mathbf{tan\,x\;=\;\frac{2t}{1\;-t^2}}$

Equation (36) can be re-written as :-

$sin\,x\;=\;2\;tan\,\frac{1}{2}x\;cos^2\,\frac{1}{2}x\;=\;\frac{2\;tan\,\frac{1}{2}x}{sec^2\,x}$

$=\;\frac{2\;tan\,\frac{1}{2}x}{1\;+\;tan^2\,\frac{1}{2}x}$

$\therefore\;\;\;\;\;\;\mathbf{sin\,x\;=\;\frac{2\,t}{1\;+\;t^2}}$

And from equation (37)

$cos\,x\;=\;cos^2\,\frac{1}{2}x(1\;-\;tan^2\,\frac{1}{2}x)\;=\;\frac{1\;-\;tan^2\,\frac{1}{2}x}{sec^2\,\frac{1}{2}x}$

$=\;\frac{1\;-\;tan^2\,\frac{1}{2}x}{1\;+\;tan^2\,\frac{1}{2}x}$

$\therefore\;\;\;\;\;\;\mathbf{cos\,x\;=\;\frac{1\;-\;t^2}{1\;+\;t^2}}$

These three equations (40); (43) ; (46) are useful in the solution of a certain type of trigonometrical equation. They also have other important applications.

### Example 1

If tan $\inline&space;\theta\;=\;\frac{4}{3}\;and\;if\;0^0\;<\;\theta\;<\;360^0$ find without tables the possible values of $\inline&space;\,tan\,\frac{1}{2}\;\theta\;and\;of\;sin\,\frac{1}{2}\;\theta$

$Let\;t\;=\;tan\,\frac{1}{2}\theta&space;\;then\;\frac{4}{3}\;=\;tan\,\theta&space;\;=\;\frac{2t}{1\;-\;t^2}$

$\therefore\;\;\;\;\;\;\;4\;-\;4t^2\;=\;6t$

$or\;\;\;\;\;\;2t^2\;+\;3t\;-\;2\;=\;0$

$\mathbf{t\;=\;\frac{1}{2}\;\;\;\;or\;\;\;\;-\,2}$

to find $\inline&space;&space;sin\,\frac{1}{2}\,\theta$

$t\;=\;tan\,\frac{1}{2}\theta&space;\;=\;sin\,\frac{1}{2}\,\theta&space;\;sec\,\frac{1}{2}\,\theta$

$=\;sin\,\frac{1}{2}\,\theta&space;(1\;+\;tan^2\,\frac{1}{2}\,\theta)^{\frac{1}{2}}$
$\therefore\;\;\;\;\;\;sin\,\frac{1}{2}\theta&space;\;=\;\frac{t}{\sqrt{1\;+\;t^2}}$

$If\;\;\;t\;=\;\frac{1}{2}\;\;\;\;then\;\;\;\;sin\,\frac{1}{2}\,\theta&space;\;=\;\frac{1}{\sqrt{5}}$

$If\;t\;=\;-\;2\;\;\then\;\;\;sin\,\frac{1}{2}\,\theta&space;\;=\;\frac{-2}{\pm&space;\sqrt{5}}\;=\;\frac{2}{\sqrt{5}}\;if\;\theta\;&space;is\;<\;&space;360^0\;and\;\frac{\theta&space;}{2}<\;180^0$

## The Auxiliary Angle

The equation $\inline&space;&space;a\;cos\,\theta\;+\;b\;sin\,\theta\;=\;c$ in which a; b; c are known numerical quantities . A method of solution is to divide throughout by $\inline&space;\sqrt{(a^2\;+\;b^2)}$

$\therefore\;\;\;\;\;\;\;\frac{a}{\sqrt{(a^2\;+\;b^2)}}cos\,\theta&space;\;+\;\frac{b}{\sqrt{(a^2\;+\;b^2)}}sin\;\theta\;=\;\frac{c}{\sqrt{a^2\;+\;b^2}}$

If we introduce an angle $\inline&space;\lambda$ whose tangent is b/a it can be seen that we can read off values for bot the sine and cosine. Hence the equation can be re-written as:-

$cos\,\theta&space;\;cos\,\lambda&space;\;+\;sin\,\theta&space;\;sin\,\lambda&space;\;=\;\frac{c}{\sqrt{(a^2\;+\;b^2)}}$

$\therefore\;\;\;\;\;\;cos\,(\theta&space;\;-\;\lambda&space;)=\;\frac{c}{\sqrt{(a^2\;+\;b^2)}}$

$cos\,\theta&space;\;cos\,\lambda&space;\;+\;sin\,\theta&space;\;sin\,\lambda&space;\;=\;\frac{c}{\sqrt{(a^2\;+\;b^2)}}$

The equation has now been reduced to one of the standard forms whose solution is known. Hence a value for $\inline&space;\theta\;-\;\lambda$ can be found and as the value of $\inline&space;\lambda$ is known $\inline&space;\theta$ ca be calculated. For real solutions it is necessary for the value of c to be less than $\inline&space;\sqrt{(a^2\;+\;b^2)}$

A second method of solution is to use the half angle formulae ( Equations (43) and (46)

$Hence\;\;\;\;\;\;a(1\;-\;t^2)\;+\;b(2t)\;=\;c(1\;+\;t^2)$

$\therefore\;\;\;\;\;\;(a\;+\;c)t^2\;-2bt\;-\;(a\;-\;c)\;=\;0$

This quadratic gives two values for t from which general value of $\inline&space;\theta$ can be found.

## The Inverse Notation

If sin$\inline&space;\theta$ = x where x is a given quantity numerically less than unity, wwe know that $\inline&space;\theta$ can be any one of a whole series of angles. Thus if $\inline&space;sin\,\theta\;=\;\frac{1}{2}\;,\;then,\theta\;=\;n\pi&space;\;+\;(-\.1)^n(\frac{\pi&space;}{6})\;and\;\theta$ can have a number of values. The inverse notation $\inline&space;\theta\;=\;sin^{-1}\.x$ is used to denote the angle whose sine is x and the numerically smallest angle satisfying the relationship $\inline&space;x\;=\;sin\,\theta$ is chosen as the principle value. Here and in what follows we shall deal only with principle values and the statement $\inline&space;\theta&space;\;=\;sin^{-1}\,x$ to mean that $\inline&space;\theta$ is the angle that lies between $\inline&space;-\frac{\pi&space;}{2}\;and\;\frac{\pi&space;}{2}$ radians whose sine is x. The statement $\inline&space;\mathbf{\theta\;=\;sin^{-1}\,x}$ means that $\inline&space;\theta$ is the inverse sine of x. On the continent this is sometimes written as $\inline&space;\mathbf{\theta&space;\;=\;arc\;sin\,x}$

The graph of $\inline&space;\theta\;=\;sin^{-1}\,x$ is, on thus that part of the graph $\inline&space;x\;=\;sin\,\theta\;given\;by\;-\;\frac{\pi}{2}\;&space;<\;\theta\;<\;\frac{\pi}{2}$ with the x-axis horizontal and the $\inline&space;\theta$ axis vertical.As shown:-

In a similar way $\inline&space;&space;\theta\;=\;cos^{-1}\,x$ will be taken to denote the smallest angle whose cosine takes the same value for negative as for positive angles and we require a notation which gives an unique value of $\inline&space;\theta$ when x is given, we conventionally take $\inline&space;\theta$ as the angle lying between 0 and $\inline&space;\pi$ radians whose cosine is x.

For example
$cos^{-1}\,\left(\frac{1}{2}&space;\right)\;=\;\frac{\pi&space;}{3}\;\;\;and\;\;\;cos^{-1}\,\left(-\,\frac{1}{2}&space;\right)\;=\;\frac{2\,\pi&space;}{3}$

The graph of $\inline&space;\theta\;=\;cos^{-1}\;x&space;\;is&space;\;derived&space;\;from\;&space;that&space;\;of&space;\;x\;=\;cos\,\theta.$

The inverse tangent is similarly defined but as, unlike the sine and cosine, the tangent can take all values, x is quite unrestricted in value. $\inline&space;\theta\;=\;tan^{-1}\;x$ is taken to mean$\inline&space;tan^{-1}(1)\;=\;\frac{\pi&space;}{4}\;\;\;and\;\;\;tan^{-1}\;(-\,1)\;=\;-\frac{\pi}{4}$n that $\inline&space;\theta$ lies between $\inline&space;&space;\frac{-\,\pi}{2}\;\;\;and\;\;\;\frac{\pi}{2}$ radians.

$\inline&space;tan^{-1}(1)\;=\;\frac{\pi&space;}{4}\;\;\;and\;\;\;tan^{-1}\;(-\,1)\;=\;-\frac{\pi}{4}$

It follows from these definitions that:-

$sin\,(sin^{-1}\,x)\;=\;x\;\;\;\;\;cos(cos^{-1}\;x)\;=\;x\;\;\;\;\;tan(tan^{-1}\,x)\;=\;x$

These relationships will be found useful in some situations.

NOTE care must be taken avoid confusion between the inverse sine, cosine etc and the reciprocal of sin x, cos x etc. The latter should always be written as :-
$\frac{1}{sin\,x}\;\;or\;\;cosec\,x\;\;\;\;and&space;\;\;\;\frac{1}{cos\,x}\;\;or\;\;sec\,x,\;\;\;etc.$