6. Trigonometrical Ratios of Angle and Sign

6.1. Values of Trigonometrical Ratios for Some Useful Angles

6.1.1. Angle of \(45^\circ\)

45 degree angle

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 \(50^\circ\)

30 degree angle

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 \(0^\circ\)

0 degree angle

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

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

supplemntary 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 \(-\theta\)

negative angled

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