6. Trigonometrical Ratios of Angle and Sign#

6.1. Values of Trigonometrical Ratios for Some Useful Angles#

6.1.1. Angle of $45^\circ$#

Consider the above figure, which is a right-angle triangle, drawn so that $\angle OMP = 90^\circ$ and $\angle MOP = 45^\circ.$ We know that sum of all angles of a triangle is equal to $180^\circ.$ Thus,

$\angle OPM = 180^\circ - \angle MOP - \angle OMP = 180^\circ - 90^\circ - 45^\circ = 45^\circ$

$\therefore OM = MP.$ Let $OP = 2a,$ then from Pythogoras theorem, we can write

$4a^2 = OP^2 = OM^2 + MP^2 = 2.OM^2\Rightarrow OM = a\sqrt{2} = MP$

$\sin 45^\circ = \frac{MP}{OP} = \frac{a\sqrt{2}}{2a} = \frac{1}{\sqrt{2}}$

Other trigonometric ratios can be deduced easily from this.

6.1.2. Angle of $30^\circ$ and $50^\circ$#

Consider an equilateral $\triangle OMP$. Let the sides $OM, OP, MP$ be each $2a$. We draw a bisector of $\angle MOP$ which will be a perpendicular bisector of $MP$ at $X$ because the triangle is equilateral. Thus, $MX = a.$ Thus, in $\triangle OMX, OM = 2a, MX = a, \angle MOX = 30^\circ, \angle OXM = 90^\circ$ because each angle is $60^\circ$ in an equilateral triangle.

$\sin MOX = \frac{MX}{OM} = \frac{1}{2} \Rightarrow \sin 30^\circ = \frac{1}{2}$

Similarly, $\angle OMX = 60^\circ$ because sum of all angles of a triangle is $180^\circ$

$\cos OMX = \frac{MX}{OM} = \frac{1}{2}\Rightarrow \cos 60^\circ = \frac{1}{2}$

All other trigonometric ratios for these can be foun from these two.

6.1.3. Angle of $0^\circ$#

Consider the $\triangle MOP$ such that the side MP is smaller than any quantiry we can assign i.e. what we denote by 0. Thus, $\angle MOP$ is what is called approaching $0$ or $\lim_{x \to 0}$ in terms of Calculus. Why we have to take such a value is because if any anngle of a triangle is equal to $0^\circ$ then the triagle won’t exist. Thus these values are limiting values as you will learn in Calculus.

However, in this case $\sin 0^\circ = \frac{MP}{OP} = \frac{0}{OP} = 0.$ Now other trigonometric ratios can be found with ease.

6.1.4. Angle of $90^\circ$#

In the previous figure as $\angle OMP$ will approach $0^\circ,$ the $\angle OPM$ will approach $90^\circ.$ Also, $OP$ will approach the length of $OM.$ Similar to previos case in right-angle triangle if one angle(other than right angle) approaches $0^\circ$ the other one will approach $90^\circ$ and at that value the triangle will cease to exist.

Thus, $\sin 90^\circ = \frac{OM}{OP} = \frac{OP}{OP} = 1.$ Now it is trivial to find other trigonometric ratios.

Given below is a table of most useful angles:

Values of useful angles#

Angle

$0^\circ$

$30^\circ$

$45^\circ$

$60^\circ$

$90^\circ$

$\sin$

$0$

$\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2}$

$1$

$\cos$

$1$

$\frac{\sqrt{3}}{2}$

$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}$

$0$

$\tan$

$0$

$\frac{1}{\sqrt{3}}$

$1$

$\sqrt{3}$

$\infty$

$\cosec$

$\infty$

$2$

$\sqrt{2}$

$\frac{2}{\sqrt{3}}$

$1$

$\sec$

$1$

$\frac{2}{\sqrt{3}}$

$\sqrt{2}$

$2$

$\infty$

$\cot$

$\infty$

$\sqrt{3}$

$1$

$\frac{1}{\sqrt{3}}$

$0$

6.2. Complementary Angles#

Angles are said to be complementary if their sum is equal to one right angle i.e $90^\circ.$ Thus, if measure of one angle is $\theta$ the other will automatically be $90^\circ - \theta.$

Consider the above figure. $\triangle OMP$ is a right-angle triangle whose $\angle OMP$ is a right angle. Since the sum of all angles is equal to $180^\circ,$ therefore sum of $\angle MOP$ and $\angle MPO$ will be equal to one right angle or $90^\circ$ i.e. they are complementatry angles.

Let $\angle MOP = \theta$ then $\angle MPO = 90^\circ - \theta.$ When $\angle MPO$ is considered $MP$ becomes the base and $OM$ becomes the perpendicular.

Thus, $\sin(90^\circ - \theta) = \sin MPO = \frac{MO}{PO} = \cos MOP = \cos \theta$

$\cos(90^\circ - \theta) = \sin MPO = \frac{PM}{PM} = \sin MOP = \sin \theta$

$\tan(90^\circ - \theta) = \tan MPO = \frac{OM}{MP} = \cot MOP = \cot \theta$

Similarly, $\cot(90^\circ - \theta) = \tan \theta$

$\cosec(90^\circ - \theta) = \sec \theta$

$\sec(90^\circ - \theta) = \cosec \theta$

6.3. Supplementary Angles#

Angles are said to be supplementary if their sum is equal to two right angles i.e. $180^\circ.$ Thus, if measure of one angle is $\theta$ and oher will automaticaly be $180^\circ - \theta.$

Conside the above figure which include the angles of $180^\circ + \theta.$ In each figure $OM$ and $OM'$ are drawn in different directions, while $MP$ and $M'P'$ are drawn in the same direction, so that

$OM' = -OM$ and $M'P' = MP$

Hencem we can say that

$\sin(180^\circ - \theta) = \sin MOP' = \frac{M'P'}{OP'} = \frac{MP}{OP} = \sin \theta$

$\cos(180^\circ - \theta) = \cos MOP' = \frac{OM'}{OP'} = \frac{-OM}{OP} = -\cos \theta$

$\tan(180^\circ - \theta) = \tan MOP' = \frac{OM'}{M'P'} = \frac{-OM}{MP} = -\tan \theta$

Similarly, $\cot(180^\circ - \theta) = -\cot \theta$

$\sec(180^\circ - \theta) = -\sec \theta$

$\cosec(180^\circ - \theta) = \cosec \theta$

6.4. Angles of $-\theta$#

Consider the above diagram which plots angles of $\theta$ and $-\theta.$ Note that $MP$ and $MP'$ are equal in magnitude but are opposite in sign. Thus, we have

$\sin(-\theta) = \frac{MP'}{OP'} = \frac{-MP}{OP} = -\sin\theta$

$\cos(-\theta) = \frac{OM}{MP'} = \frac{OM}{OP} = \cos\theta$

$\tan(-\theta) = \frac{MP'}{OM} = \frac{-MP}{OM} = -\tan\theta$

Similalry, $\cot(-\theta) = -\cot\theta$

$\sec(-\theta) = \sec\theta$

$\cosec(-\theta) = -\cosec\theta$

6.5. Angles of $90^\circ -\theta$ and $90^\circ+\theta$#

Similarly it can be proven that(diagram has been left as an exercise)

$\sin(90^\circ -\theta) = \cos\theta$

$\cos(90^\circ -\theta) = \sin\theta$

$\tan(90^\circ -\theta) = \cot\theta$

$\cot(90^\circ -\theta) = \tan\theta$

$\sec(90^\circ -\theta) = \cosec\theta$

$\cosec(90^\circ -\theta) = \sec\theta$

$\sin(90^\circ+\theta) = \cos\theta$

$\cos(90^\circ+\theta) = -\sin\theta$

$\tan(90^\circ+\theta) = -\cot\theta$

$\cot(90^\circ+\theta) = -\tan\theta$

$\sec(90^\circ+\theta) = -\cosec\theta$

$\cosec(90^\circ+\theta) = \sec\theta$

6.6. Angles of $180^\circ + \theta$#

Angles of $180^\circ + \theta, 270^\circ -\theta, 270^\circ + theta$ can be found by using previous relations. For example,

$\sin(180 + \theta) = \sin(90 + 90 + \theta) = \cos(90 + \theta) = -\sin\theta$

$\cos(180 + \theta) = \cos(90 + 90 + \theta) = -\sin(90 + \theta) = -\cos\theta$

$\tan(180 + \theta) = \tan(90 + 90 + \theta) = -\cos(90 + \theta) = \tan\theta$

Similarly, $\cot(180 + \theta) = \cot\theta$

$\sec(180 + \theta) = -\sec\theta$

$\cosec(180 + \theta) = -\cosec\theta$

6.7. Angles of $360^\circ + \theta$#

For angles of $\theta$ the radius vector makes an angle of $\theta$ with initial side. For angles of $360^\circ + \theta$ it will complete a full revolution and then make an angle of $\theta$ with initial side. Thus, the trigonometrical ratios for an angle of $360^\circ + \theta$ are the same as those for $\theta.$

It is clear that angle will remain $\theta$ for any multiple of $360^\circ.$

6.8. Problems#

1. If $A = 30^\circ,$ verify that

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

2. $\sin 2A = 2\sin A\cos A$

3. $\cos 3A = 4\cos^3A - 3\cos A$

4. $\sin 3A = 3\sin A - 4\sin^3A$

5. $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

2. If $A = 45^\circ,$ verify that

1. $\sin 2A = 2\sin A\cos A$

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

3. $\tan 2A = \frac{2\tan A}{1 - \tan^2A}$

Verify that

1. $\sin^230^\circ + \sin^245^\circ + \sin^260^\circ = \frac{3}{2}$

2. $\tan^230^\circ + \tan^245^\circ + \tan^260^\circ = 4\frac{1}{3}$

3. $\sin 30^\circ\cos 60^\circ + \sin 60^\circ\cos 30^\circ = 1$

4. $\cos 45^\circ\cos 60^\circ - \sin 45^\circ\sin 60^\circ = -\frac{\sqrt{3} - 1}{2\sqrt{2}}$

5. $\cosec^245^\circ.\sec^230^\circ.\sin^290^\circ.\cos 60^\circ = 1\frac{1}{3}$

6. $4\cot^245^\circ-\sec^260^\circ + \sin^230^\circ = \frac{1}{4}$

Prove that

1. $\sin 420^\circ\cos 390^\circ + \cos(-300^\circ)\sin(-330^\circ) = 1$

2. $\cos 570^\circ\sin 510^\circ -\sin 330^\circ\cos 390^\circ = 0$

What are the values of $\cos A - \sin A$ and $\tan A + \cot A$ when A has the values

1. $\frac{\pi}{3}$

2. $\frac{2\pi}{3}$

3. $\frac{5\pi}{4}$

4. $\frac{7\pi}{4}$

5. $\frac{11\pi}{3}$

What values between $0^\circ$ and $360^\circ$ may $A$ have when

1. $\sin A = \frac{1}{\sqrt{2}}$

2. $\cos A = -\frac{1}{2}$

3. $\tan A = -1$

4. $\cot A = -\sqrt{3}$

5. $\sec A = -\frac{2}{\sqrt{3}}$

6. $\cosec A = -2$

Express in terms of the ratios of a positive angle, which is less than $45^\circ,$ the quantities

1. $\sin(-65^\circ)$

2. $\cos(-84^\circ)$

3. $\tan 137^\circ$

4. $\sin 168^\circ$

5. $\cos 287^\circ$

6. $\tan(-246^\circ)$

7. $\sin 843^\circ$

8. $\cos(-928^\circ)$

9. $\tan 1145^\circ$

10. $\cos 1410^\circ$

11. $\cot(-1054^\circ)$

12. $\sec 1327^\circ$

13. $\cosec (-756^\circ)$

What sign has $\sin A + \cos A$ for the following values of $A$?

1. $140^\circ$

2. $278^\circ$

3. $-356^\circ$

4. $-1125^\circ$

What sign has $\sin A - \cos A$ for the following values of $A$?

1. $215^\circ$

2. $825^\circ$

3. $-634^\circ$

4. $-457^\circ$

5. Find the sine and cosine of all angles in the first four quadrants whose tangents are equal to $\cos 135^\circ.$

Prove that

1. $\sin(270^\circ + A) = -\cos A$ and $\tan(270^\circ + A) = -\cot A$

2. $\cos(270^\circ - A) = -\sin A$ and $\cot(270^\circ - A) = \tan A$

3. $\cos A + \sin(270^\circ + A) - \sin(270^\circ - A) + \cos(180^\circ + A) = 0$

4. $\sec(270^\circ - A)\sec(90^\circ - A) - \tan(270^\circ - A)\tan(90^\circ + A) + 1 = 0$

5. $\cot A + \tan(180^\circ + A) + \tan(90^\circ + A) + \tan(360^\circ - A) = 0$

6. Find the value of $3\tan^245^\circ - \sin^260^\circ - \frac{1}{2}\cot^230^\circ + \frac{1}{8}\sec^245^\circ$

7. Simplify $\frac{\sin 300^\circ.\tan 330^\circ.\sec 420^\circ}{\tan 135^\circ.\sin 210^\circ.\sec 315^\circ}$

8. Show that $\tan 1^\circ\tan 2^\circ \ldots \tan 89^\circ = 1$

9. Show that $\sin^25^\circ + \sin^210^\circ + \sin^215^\circ + \ldots + \sin^290^\circ = 9\frac{1}{2}$

10. Find the value of $\cos^2\frac{\pi}{16} + \cos^2\frac{3\pi}{16} + \cos^2\frac{5\pi}{16} + \cos^2\frac{7\pi}{16}$

Find the value of the following:

1. $\sec^2\frac{\pi}{6}\sec^2\frac{\pi}{4} + \tan^2\frac{\pi}{3}\sin^2\frac{\pi}{2}$

2. $\cot^230^\circ - 2\cos^260^\circ - \frac{3}{4}\sin^245^\circ - 4\sin^230^\circ$

3. $\frac{\sec 480^\circ\cosec 570^\circ.\tan 330^\circ}{\sin 600^\circ.\cos 660^\circ.\cot 405^\circ}$

4. If $A = 30^\circ,$ show that $\cos^6A + \sin^6A = 1 - \sin^2A\cos^2A$

5. Show that $\left(\tan \frac{\pi}{4} + \cot \frac{\pi}{4} + \sec\frac{\pi}{4}\right)\left(\tan \frac{\pi}{4} + \cot \frac{\pi}{4} - \sec \frac{\pi}{4}\right) = \cosec^2 \frac{\pi}{4}$

6. Show that $\sin^26^\circ + \sin6^212^\circ + \sin^218^\circ + \ldots + \sin^284^\circ + \sin^290^\circ = 8$

7. Show that $\tan 9^\circ.\tan 27^\circ.\tan 45^\circ.\tan 63^\circ.\tan 81^\circ = 1$

8. Show that $\sum_{r = 1}^9 \sin^2\frac{r\pi}{18} = 5$

9. If $4n\alpha = \pi,$ show that $\tan\alpha\tan2\alpha\tan3\alpha. \ldots .\tan(2n - 2)\alpha\tan(2n - 1)\alpha = 1$