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 $$\tan\alpha\tan2\alpha\tan3\alpha. \ldots .\tan(2n - 2)\alpha\tan(2n - 1)\alpha = 1$$