# 3.12 Trigonometric values, equations and identities  (Page 3/3)

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$⇒x=2\pi -\theta =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

Problem : Find angles in [0,2π], if

$\mathrm{cot}x=\frac{1}{\sqrt{3}}$

Solution : Considering only the magnitude of numerical value, we have :

$⇒\mathrm{cot}\theta =\frac{1}{\sqrt{3}}=\mathrm{cot}\frac{\pi }{3}$

Thus, required acute angle is π/3. Now, cotangent function is positive in first and third quadrants. Looking at the value diagram, the angle in third quadrant is :

$⇒x=\pi +\theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$

Hence angles are π/3 and 4π/3.

## Negative angles

When we consider angle as a real number entity, we need to express angles as negative angles as well. The corresponding negative angle (y) is obtained as :

$y=x-2\pi$

Thus, negative angles corresponding to 4π/3 and 5π/3 are :

$⇒y=\frac{4\pi }{3}-2\pi =-\frac{2\pi }{3}$ $⇒y=\frac{5\pi }{3}-2\pi =-\frac{\pi }{3}$

We can also find negative angle values using a separate negative value diagram (see figure). We draw negative value diagram by demarking quadrants with corresponding angles and writing angle values for negative values. We deduct “2π” from the relation for positive value diagram.

Let us consider sinx = -√3/2 again. The acute angle in first quadrant is π/3. Sine is negative in third and fourth quadrants. The angles in these quadrants are :

$y=-\pi +\theta =-\pi +\frac{\pi }{3}=-\frac{2\pi }{3}$ $y=-\theta =-\frac{\pi }{3}$

## Zeroes of sine and cosine functions

Trigonometric equations are formed by equating trigonometric functions to zero. The solutions of these equations are :

1 : $\mathrm{sin}x=0\phantom{\rule{1em}{0ex}}⇒x=n\pi ;n\in Z$

2 : $\mathrm{cos}x=0\phantom{\rule{1em}{0ex}}⇒x=\left(2n+1\right)\frac{\pi }{2};n\in Z$

## Definition of other trigonometric functions

We define other trigonometric functions in the light of zeroes of sine and cosine as listed above :

$\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x};x\ne \left(2n+1\right)\frac{\pi }{2};n\in Z$ $\mathrm{cot}x=\frac{\mathrm{cos}x}{\mathrm{sin}x};x\ne n\pi ;n\in Z$ $\mathrm{cosec}x=\frac{1}{\mathrm{sin}x};x\ne n\pi ;n\in Z$ $\mathrm{sec}x=\frac{1}{\mathrm{cos}x};x\ne \left(2n+1\right)\frac{\pi }{2};n\in Z$

## Trigonometric equations

Trigonometric function can be used to any other values as well. Solutions of such equations are given here without deduction for reference purpose. Solutions of three equations involving sine, cosine and tangent functions are listed here :

1. Sine equation

$\mathrm{sin}x=a=\mathrm{sin}y$

$x=n\pi +{\left(-1\right)}^{n}y;n\in Z$

2. Cosine equation

$\mathrm{cos}x=a=\mathrm{cos}y$

$x=2n\pi ±y;n\in Z$

3. Tangent equation

$\mathrm{tan}x=a=\mathrm{tan}y$

$x=n\pi +y;n\in Z$

In order to understand the working with trigonometric equation, let us consider an equation :

$\mathrm{sin}x=-\frac{\sqrt{3}}{2}$

As worked out earlier, -√3/2 is sine value of two angles in the interval [0, π]. Important question here is to know which angle should be used in the solution set. Here,

$⇒\mathrm{sin}\frac{4\pi }{3}=\mathrm{sin}\frac{5\pi }{3}=-\frac{\sqrt{3}}{2}$

We can write general solution using either of two values.

$x=n\pi +{\left(-1\right)}^{n}\frac{4\pi }{3};n\in Z$ $⇒x=n\pi +{\left(-1\right)}^{n}\frac{5\pi }{3};n\in Z$

The solution sets appear to be different, but are same on expansion. Conventionally, however, we use the smaller of two angles which lie in the interval [0, π]. In order to check that two series are indeed same, let us expand series from n=-4 to n=4,

For $⇒x=n\pi +{\left(-1\right)}^{n}\frac{4\pi }{3};n\in Z$

$-4\pi +\frac{4\pi }{3}=-\frac{8\pi }{3},-3\pi -\frac{4\pi }{3}=-\frac{13\pi }{3},-2\pi +\frac{4\pi }{3}=-\frac{2\pi }{3},-\pi -\frac{4\pi }{3}=-\frac{7\pi }{3},$

$0+4\pi /3=\frac{4\pi }{3},\pi -\frac{4\pi }{3}=-\frac{\pi }{3},2\pi +\frac{4\pi }{3}=\frac{10\pi }{3},3\pi -\frac{4\pi }{3}=\frac{5\pi }{3},4\pi +\frac{4\pi }{3}=\frac{16\pi }{3}$

Arranging in increasing order :

$-\frac{13\pi }{3},-\frac{8\pi }{3},-\frac{7\pi }{3},-\frac{2\pi }{3},-\frac{\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},\frac{10\pi }{3},\frac{16\pi }{3}$

For $⇒x=n\pi +{\left(-1\right)}^{n}\frac{5\pi }{3};n\in Z$

$-4\pi +\frac{5\pi }{3}=-\frac{7\pi }{3},-3\pi -\frac{5\pi }{3}=-\frac{14\pi }{3},-2\pi +\frac{5\pi }{3}=-\frac{\pi }{3},-\pi -\frac{5\pi }{3}=-\frac{8\pi }{3},$

$0+\frac{5\pi }{3}=\frac{5\pi }{3},\pi -\frac{5\pi }{3}=-\frac{2\pi }{3},2\pi +\frac{5\pi }{3}=\frac{11\pi }{3},3\pi -\frac{5\pi }{3}=\frac{4\pi }{3},4\pi +\frac{5\pi }{3}=\frac{17\pi }{3}$

Arranging in increasing order :

$-\frac{14\pi }{3},-\frac{8\pi }{3},-\frac{7\pi }{3},-\frac{2\pi }{3},-\frac{\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},\frac{11\pi }{3},\frac{17\pi }{3}$

We see that there are common terms. There are, however, certain terms which do not appear in other series. We can though find those missing terms by evaluating some more values. For example, if we put n = 6 in the second series, then we get the missing term -13π/3. Also, putting n=5,7, we get 10π/3 and 16π/3. Thus, all missing terms in second series are obtained. Similarly, we can compute few more values in first series to find missing terms. We, therefore, conclude that both these series are equal.

Problem : Find solution of equation :

$2{\mathrm{cos}}^{2}x+3\mathrm{sin}x=0$

Solution : Our objective here is to covert equation to linear form. Here, we can not convert sine term to cosine term, but we can convert ${\mathrm{cos}}^{2}x$ in terms of ${\mathrm{sin}}^{2}x$ .

$⇒2\left(1-\mathrm{sin}{}^{2}x\right)+3\mathrm{sin}x=0$ $⇒2-2\mathrm{sin}{}^{2}x+3\mathrm{sin}x=0$ $⇒2\mathrm{sin}{}^{2}x-3\mathrm{sin}x-2=0$

It is a quadratic equation in sinx. Factoring, we have :

$⇒2\mathrm{sin}{}^{2}x+\mathrm{sin}x-4\mathrm{sin}x-2=0$ $⇒\mathrm{sin}x\left(2\mathrm{sin}x+1\right)-2\left(2\mathrm{sin}x+1\right)=0$ $⇒\left(2\mathrm{sin}x+1\right)\left(\mathrm{sin}x-2\right)=0$

Either, sinx=-1/2 or sinx = 2. But sinx can not be equal to 2. hence,

$⇒\mathrm{sin}x=-\frac{1}{2}=\mathrm{sin}\left(\pi +\frac{\pi }{6}\right)=\mathrm{sin}\left(\frac{7\pi }{6}\right)$ $⇒x=n\pi +{\left(-1\right)}^{n}\frac{7\pi }{6};\phantom{\rule{1em}{0ex}}n\in Z$

Note : We shall not work with any other examples here as purpose of this module is only to introduce general concepts of angles, identities and equations. These topics are part of separate detailed study.

## Reciprocal identities

Reciprocals are defined for values of x for which trigonometric function in the denominator is not zero.

$\mathrm{sin}x=\frac{1}{\mathrm{cosec}x};\phantom{\rule{1em}{0ex}}\mathrm{cos}x=\frac{1}{\mathrm{sec}x};\phantom{\rule{1em}{0ex}}\mathrm{tan}x=\frac{1}{\mathrm{cot}x};$ $\mathrm{cosec}x=\frac{1}{\mathrm{sin}x};\phantom{\rule{1em}{0ex}}\mathrm{sec}x=\frac{1}{\mathrm{cos}x};\phantom{\rule{1em}{0ex}}\mathrm{cot}x=\frac{1}{\mathrm{tan}x}$

## Negative angle identities

$\mathrm{cos}\left(-x\right)=\mathrm{cos}x;\phantom{\rule{1em}{0ex}}\mathrm{sin}\left(-x\right)=-\mathrm{sin}x;\phantom{\rule{1em}{0ex}}\mathrm{tan}\left(-x\right)=-\mathrm{tan}x$

## Pythagorean identities

${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1;\phantom{\rule{1em}{0ex}}1+{\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x;\phantom{\rule{1em}{0ex}}1+{\mathrm{cot}}^{2}x={\mathrm{cosec}}^{2}x$

## Sum/difference identities

$\mathrm{sin}\left(x±y\right)=\mathrm{sin}x\mathrm{cos}y±\mathrm{sin}y\mathrm{cos}x$ $\mathrm{cos}\left(x±y\right)=\mathrm{cos}x\mathrm{cos}y\mp \mathrm{sin}x\mathrm{sin}y$ $\mathrm{tan}\left(x±y\right)=\mathrm{tan}sx±\mathrm{tan}y/1\mp \mathrm{tan}x\mathrm{tan}y;\phantom{\rule{1em}{0ex}}\text{x,y and (x+y) are not odd multiple of π/2}$ $\mathrm{cot}\left(x±y\right)=\mathrm{cot}x\mathrm{cot}y\mp 1/\mathrm{cot}y±\mathrm{cot}x;\phantom{\rule{1em}{0ex}}\text{x,y and (x+y) are not odd multiple of π/2}$

## Double angle identities

$\mathrm{sin}2x=2\mathrm{sin}x\mathrm{cos}x=\frac{2\mathrm{tan}x}{1+\mathrm{tan}{}^{2}x}$ $\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=2{\mathrm{cos}}^{2}x-1=1-2{\mathrm{sin}}^{2}x=\frac{1-{\mathrm{tan}}^{2}x}{1+{\mathrm{tan}}^{2}x}$ $\mathrm{tan}2x=\frac{2\mathrm{tan}x}{1-\mathrm{tan}{}^{2}x}$ $\mathrm{cot}2x=\frac{{\mathrm{cot}}^{2}x-1}{2\mathrm{cot}x}$

## Triple angle identities

$\mathrm{sin}3x=3\mathrm{sin}x-4{\mathrm{sin}}^{3}x$ $\mathrm{cos}3x=4{\mathrm{cos}}^{3}x-3\mathrm{cos}x$ $\mathrm{tan}3x=\frac{3\mathrm{tan}x-{\mathrm{tan}}^{3}x}{1-3{\mathrm{tan}}^{2}x}$ $\mathrm{cot}3x=\frac{3\mathrm{cot}x-{\mathrm{cot}}^{3}x}{1-3{\mathrm{cot}}^{2}x}$

## Power reduction identities

$\mathrm{sin}{}^{2}x=\frac{1-\mathrm{cos}2x}{2}$ $\mathrm{cos}{}^{2}x=\frac{1+\mathrm{cos}2x}{2}$ $\mathrm{sin}{}^{3}x=\frac{3\mathrm{sin}x-\mathrm{sin}3x}{4}$ ${\mathrm{cos}}^{3}x=\frac{\mathrm{cos}3x+3\mathrm{cos}x}{4}$

## Product to sum identities

$2\mathrm{sin}x\mathrm{cos}y=\mathrm{sin}\left(x+y\right)+\mathrm{sin}\left(x-y\right)$ $2\mathrm{cos}x\mathrm{sin}y=\mathrm{sin}\left(x+y\right)-\mathrm{sin}\left(x-y\right)$ $2\mathrm{cos}x\mathrm{cos}y=\mathrm{cos}\left(x+y\right)+\mathrm{cos}\left(x-y\right)$ $2\mathrm{sin}x\mathrm{sin}y=-\mathrm{cos}\left(x+y\right)+\mathrm{cos}\left(x-y\right)=\mathrm{cos}\left(x-y\right)-\mathrm{cos}\left(x+y\right)$

## Sum to product identities

$\mathrm{sin}x+\mathrm{sin}y=2\mathrm{sin}\frac{\left(x+y\right)}{2}\mathrm{cos}\frac{\left(x-y\right)}{2}$ $\mathrm{sin}x-\mathrm{sin}y=2\mathrm{cos}\frac{\left(x+y\right)}{2}\mathrm{sin}\frac{\left(x-y\right)}{2}$ $\mathrm{cos}x+\mathrm{cos}y=2\mathrm{cos}\frac{\left(x+y\right)}{2}\mathrm{cos}\frac{\left(x-y\right)}{2}$ $\mathrm{cos}x-\mathrm{cos}y=-2\mathrm{sin}\frac{\left(x+y\right)}{2}\mathrm{sin}\frac{\left(x-y\right)}{2}=2\mathrm{sin}\frac{\left(x+y\right)}{2}\mathrm{sin}\frac{\left(y-x\right)}{2}$

## Half angle identities

$\mathrm{sin}\frac{x}{2}=±\sqrt{\left\{\frac{\left(1-\mathrm{cos}x\right)}{2}\right\}}$ $\mathrm{cos}\frac{x}{2}=±\sqrt{\left\{\frac{\left(1+\mathrm{cos}x\right)}{2}\right\}}$ $\mathrm{tan}\frac{x}{2}=\mathrm{cosec}x-\mathrm{cot}x=±\sqrt{\left\{\frac{\left(1-\mathrm{cos}x\right)}{\left(1+\mathrm{cos}x\right)}\right\}}=\frac{\mathrm{sin}x}{1+\mathrm{cos}x}=\frac{1-\mathrm{cos}x}{\mathrm{sin}x}$ $\mathrm{cot}\frac{x}{2}=\mathrm{cosec}x+\mathrm{cot}x=±\sqrt{\left\{\frac{\left(1+\mathrm{cos}x\right)}{\left(1-\mathrm{cos}x\right)}\right\}}=\frac{\mathrm{sin}x}{1-\mathrm{cos}x}=\frac{1+\mathrm{cos}x}{\mathrm{sin}x}$

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