Overview: Trigonometry

Basic trigonometric functions

The trigonometric functions relate the angles of a right-angled triangle to the ratios of its sides. They are widely used in geometry, mensuration, algebraic manipulation, coordinate geometry, calculus and in many applied problems. Two standard viewpoints are used: the right-triangle definition and the unit-circle definition. Both agree for angles between 0° and 90° and extend consistently to all real angles.

• Sine (sin): In a right triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.
• Cosine (cos): In a right triangle, the cosine of an angle is the ratio of the length of the adjacent side to the hypotenuse.
• Tangent (tan): The tangent of an angle is the ratio of the opposite side to the adjacent side.
• Cotangent (cot): The cotangent is the reciprocal of tangent: cot x = 1 / tan x.
• Secant (sec): The secant is the reciprocal of cosine: sec x = 1 / cos x.
• Cosecant (csc): The cosecant is the reciprocal of sine: csc x = 1 / sin x.

Unit-circle view and signs

On the unit circle (radius = 1), for an angle measured from the positive x-axis,
\( \sin \theta = y \)
\( \cos \theta = x \)
\( \tan \theta = \dfrac{y}{x} \) provide x not equal to 0.

Signs of sin, cos, tan depend on the quadrant of the angle. Sine and cosine values for standard angles are useful to memorise:

• \( \sin 0^\circ = 0,\ \sin 30^\circ = \tfrac{1}{2},\ \sin 45^\circ = \tfrac{\sqrt{2}}{2},\ \sin 60^\circ = \tfrac{\sqrt{3}}{2},\ \sin 90^\circ = 1\).
• \( \cos 0^\circ = 1,\ \cos 30^\circ = \tfrac{\sqrt{3}}{2},\ \cos 45^\circ = \tfrac{\sqrt{2}}{2},\ \cos 60^\circ = \tfrac{1}{2},\ \cos 90^\circ = 0\).
• \( \tan 0^\circ = 0,\ \tan 30^\circ = \tfrac{1}{\sqrt{3}},\ \tan 45^\circ = 1,\ \tan 60^\circ = \sqrt{3}\).

Trigonometric identities

Identities are equalities true for all values of the variable where both sides are defined. They are used to simplify expressions and solve equations.

Pythagorean identities

\( \sin^2 x + \cos^2 x = 1 \)
\( 1 + \tan^2 x = \sec^2 x \)
\( 1 + \cot^2 x = \csc^2 x \)

Reciprocal and quotient identities

\( \csc x = \dfrac{1}{\sin x},\ \sec x = \dfrac{1}{\cos x},\ \cot x = \dfrac{1}{\tan x} \)
\( \tan x = \dfrac{\sin x}{\cos x},\ \cot x = \dfrac{\cos x}{\sin x} \)

Addition and subtraction formulas

\( \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \)
\( \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \)
\( \tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} \)
\( \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta \)
\( \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \)
\( \tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} \)

Double-angle formulas

\( \sin 2x = 2\sin x\cos x \)
\( \cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x \)
\( \tan 2x = \dfrac{2\tan x}{1 - \tan^2 x} \)

Half-angle formulas

\( \sin^2 \dfrac{x}{2} = \dfrac{1 - \cos x}{2} \)
\( \cos^2 \dfrac{x}{2} = \dfrac{1 + \cos x}{2} \)
\( \tan \dfrac{x}{2} = \dfrac{\sin x}{1 + \cos x} = \dfrac{1 - \cos x}{\sin x} \)

Other useful transformations

• Sum-to-product: \( \sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \)
• Product-to-sum: \( \sin\alpha\cos\beta = \tfrac{1}{2}[\sin(\alpha+\beta) + \sin(\alpha-\beta)] \)
• Phase shift: \( a\sin x + b\cos x = R\sin(x+\phi) \) where \( R = \sqrt{a^2+b^2} \) and \( \phi \) satisfies \( \cos\phi = a/R,\ \sin\phi = b/R \).

Trigonometric equations

Solving trigonometric equations means finding all angles (usually in a specified interval or all real solutions) that satisfy an equation involving trigonometric functions. Common methods include using known identities, algebraic manipulation, factoring, substitution and using the unit circle to find principal solutions followed by general solutions.

Standard approaches

• Use identities to convert the equation to a single function of an angle (for example, express everything in terms of sin x or tan x).
• Factor the expression, if possible, to split into simpler equations.
• Use substitution such as \( t = \tan \dfrac{x}{2} \) (Weierstrass substitution) for rationalising expressions in sin and cos.
• Use double-angle or half-angle formulas when the equation involves \(2x\) or \(x/2\).
• Find principal solutions on \([0,2\pi)\) or \([0^\circ,360^\circ)\) then add the appropriate period to get general solutions:

  \( x = \text{principal solution} + 2k\pi \) for sin and cos;

  \( x = \text{principal solution} + k\pi \) for tan and cot, where \(k\) is any integer.

Factoring and common errors

Factoring trigonometric expressions follows the same algebraic ideas as polynomial factoring. Common transformations include taking a common factor, grouping, or using product-to-sum identities. Be careful: expressions like \( \sin x + \cos x \) do not factor as a simple product of sin and cos. The correct useful identity is:

\( \sin x + \cos x = \sqrt{2}\,\sin\!\left(x + \dfrac{\pi}{4}\right) \)

Reciprocal and other special cases

Equations that involve reciprocals (for example \( \csc x = \sec x \)) are reduced to regular trigonometric equations by taking reciprocals on both sides when valid, or by converting reciprocals using the reciprocal identities. For example:

\( \dfrac{1}{\sin x} = \dfrac{1}{\cos x} \)
\( \Rightarrow \sin x = \cos x \)
\( \Rightarrow \tan x = 1 \)

Examples

Q1: A tower has a height of 50 m. It has an angle of elevation from two points on the ground level on its opposite sides are 30 degree and 60 degree respectively. Calculate the distance between two points.
(a)110.97 m
(b) 107.56 m
(c) 115.47 m
(d) 120.93 m
(e) None of the above
Ans: c
Sol: Let the vertical tower base be point O and the two ground points be A and B on opposite sides with angles of elevation \(30^\circ\) at A and \(60^\circ\) at B. Let OA = \(x_1\) and OB = \(x_2\).
\( \tan 30^\circ = \dfrac{50}{x_1} \)
\( \tan 30^\circ = \dfrac{1}{\sqrt{3}} \)
\( x_1 = \dfrac{50}{\tan 30^\circ} = 50\sqrt{3} \)
\( \tan 60^\circ = \dfrac{50}{x_2} \)
\( \tan 60^\circ = \sqrt{3} \)
\( x_2 = \dfrac{50}{\tan 60^\circ} = \dfrac{50}{\sqrt{3}} = \dfrac{50\sqrt{3}}{3} \)
The distance between the two points A and B is \( x_1 + x_2 \).
\( x_1 + x_2 = 50\sqrt{3} + \dfrac{50\sqrt{3}}{3} = \dfrac{200\sqrt{3}}{3} \)
\( \dfrac{200\sqrt{3}}{3} \approx 115.47\ \text{m} \)
Therefore the correct choice is c.

Q2: The value of 32cot2π/4 - 8sec2π/3 + 8cos3π/6
(a) √3
(b) 2√3
(c) 3
(d) 3√3
(e) None of the above/More than one of the above.
Ans: e
Sol: Evaluate each term separately.
\( 2\pi/4 = \pi/2 \)
\( \cot \dfrac{\pi}{2} = \dfrac{\cos(\pi/2)}{\sin(\pi/2)} = \dfrac{0}{1} = 0 \)
\( 32\cot\dfrac{2\pi}{4} = 32 \times 0 = 0 \)
\( 2\pi/3 = 120^\circ \)
\( \cos \dfrac{2\pi}{3} = -\dfrac{1}{2} \)
\( \sec \dfrac{2\pi}{3} = \dfrac{1}{\cos(2\pi/3)} = -2 \)
\( -8\sec\dfrac{2\pi}{3} = -8 \times (-2) = 16 \)
\( 3\pi/6 = \dfrac{\pi}{2} \)
\( \cos \dfrac{\pi}{2} = 0 \)
\( 8\cos\dfrac{3\pi}{6} = 8 \times 0 = 0 \)
Sum of terms \( = 0 + 16 + 0 = 16 \)
None of the options (a-d) equals 16. Therefore the correct choice is e (None of the above).

Concluding notes and practice tips

• Memorise values of sin, cos and tan for standard angles (0°, 30°, 45°, 60°, 90°) and their radian equivalents; these are used repeatedly in competitive problems.
• Use identities to convert mixed expressions to a single function before solving; converting to sin and cos is often helpful.
• When solving equations, always consider the domain and add the correct period to obtain general solutions.
• Draw a diagram for geometry problems involving angles of elevation and depression; translate the diagram into equations using tan, sin or cos as appropriate.
• Check special cases where denominators may become zero (for example when converting using reciprocals) to avoid extraneous solutions.