Definition of Trigonometric Functions
In a right angled triangle ABC, ∠CAB = A and ∠BCA = 90° = π/2. AC is the base, BC the altitude and AB is the hypotenuse. We refer to the base as the adjacent side and to the altitude as the opposite side. There are six trigonometric ratios, also called trigonometric functions or circular functions. With reference to angle A, the six ratios are:
BC/AB = opposite side / hypotenuse is called sine of A, and written as sin A
AC/AB = adjacent side / hypotenuse is called the cosine of A, and written as cos A
BC / AC = opposite side / adjacent side is called the tangent of A, and written as tan A.
Obviously, tan A = sin A/cos A. The reciprocals of sine, cosine and tangent are called the cosecant, secant and cotangent of A respectively. We write these as cosec A, sec A, cot A respectively.
Basic Trigonometric Formulae
BASIC FORMULAE
Ex.:
cos6 A + sin6 A + 3 sin2 A cos2 A is equal to
Solution:
(B). sin6 A + cos6 A + 3 sin2 A cos2 A
= sin6 A + cos6 A + 3 sin2 A cos2 A(sin2 A + cos2 A) = (sin2 A + cos2 A)3 = 1.
Trigonometric Ratios of Any Angle
| α equals | sin α | cos α | tan α | cot α | Sec α | cosec α |
|---|---|---|---|---|---|---|
| -θ | -sin θ | cos θ | -tan θ | -cot θ | sec θ | -cosec θ |
| 90° - θ | cos θ | sin θ | cot θ | tan θ | cosec θ | sec θ |
| 90° + θ | cos θ | -sin θ | -cot θ | -tan θ | -cosec θ | sec θ |
| 180° - θ | sin θ | -cos θ | -tan θ | -cot θ | -sec θ | cosec θ |
| 180° + θ | -sin θ | -cos θ | tan θ | cot θ | -sec θ | -cosec θ |
| 360° - θ | -sin θ | cos θ | -tan θ | -cot θ | sec θ | -cosec θ |
| 360° + θ | sin θ | cos θ | tan θ | cot θ | sec θ | cosec θ |
Notes:
- Angle θ and 90° − θ are complementary angles; θ and 180° − θ are supplementary angles
- sin(nπ + (−1)nθ) = sinθ, n ∈ ℤ
- cos(2nπ ± θ) = cosθ, n ∈ ℤ
- tan(nπ + θ) = tanθ, n ∈ ℤ
TRIGONOMETRIC RATIOS OF COMPOUND ANGLES
An angle made up of the algebraic sum of two or more angles is called a compound angle. Some of the formulae and results regarding compound angles are:
- sin(A + B) = sin A cos B + cos A sin B
- sin(A – B) = sin A cos B – cos A sin B
- cos(A + B) = cos A cos B – sin A sin B
- cos(A – B) = cos A cos B + sin A sin B
- tan(A + B) = (tan A + tan B) / (1 – tan A tan B)
- tan(A – B) = (tan A – tan B) / (1 + tan A tan B)
- sin(A + B) sin(A – B) = sin²A – sin²B = cos²B – cos²A
- cos(A + B) cos(A – B) = cos²A – sin²B = cos²B – sin²A
TRIGONOMETRIC RATIOS OF MULTIPLES OF AN ANGLE
- sin2A = 2sinA cosA = 2 tanA1 + tan2A
- cos2A = cos2A − sin2A = 1 − 2sin2A = 2cos2A − 1 = 1 − tan2A1 + tan2A
- tan2A = 2 tanA1 − tan2A
- sin3A = 3sinA − 4sin3A = 4sin(60° − A) sinA sin(60° + A)
- cos3A = 4cos3A − 3cosA = 4cos(60° − A) cosA cos(60° + A)
- tan3A = 3 tanA − tan3A1 − 3 tan2A = tan(60° − A) tanA tan(60° + A)
SUM OF SINES/COSINES IN TERMS OF PRODUCTS
- sinA + sinB = 2sin[(A+B)/2] cos[(A−B)/2]
- sinA − sinB = 2sin[(A−B)/2] cos[(A+B)/2]
- cosA + cosB = 2cos[(A−B)/2] cos[(A+B)/2]
- cosA − cosB = 2sin[(B−A)/2] sin[(A+B)/2] (here notice (B−A)!)
Conversely:
- 2sinAcosB = sin(A+B) + sin(A−B)
- 2cosAsinB = sin(A+B) − sin(A−B)
- 2cosAcosB = cos(A+B) + cos(A−B)
- 2sinAsinB = cos(A−B) − cos(A+B)
Example:
sin 12° · sin 48° · sin 54° is equal to
(A) 1 (B) 1/4 (C) 1/8 (D) none of these
Solution: (C).
Let θ = 12°.
L.H.S. = (1/sin 72°) sin12° sin48° sin72° sin54°
= (1/4) sin 3(12°) sin54° / sin72°
= sin36° sin54° / 8 sin36° cos36°
= cos36° / 8 cos36°
= 1/8 = R.H.S.
Trigonometric Series:
∏r=0n-1 cos 2r A = sin 2n A
2n sin A where ∏ denotes products.
Formulas and Concepts
Trigonometric Formulae
Identitities
- sin²x + cos²x = 1, for all x ∈ ℝ
- cosec²x = 1 + cot²x, x ≠ nπ
- sec²x = 1 + tan²x, x, x ≠ (2n + 1)π/2
Sum and Difference Fromula
- sin(A + B) = sinA cosB + cosA sinB
- sin(A − B) = sinA cosB − cosA sinB
- cos(A + B) = cosA cosB − sinA sinB
- cos(A − B) = cosA cosB + sinA sinB
- tan(A + B) = (tanA + tanB) / (1 − tanA tanB)
- tan(A − B) = (tanA − tanB) / (1 + tanA tanB)
- sin(A + B) sin(A − B) = sin²A − sin²B = cos²B − cos²A
- cos(A + B) cos(A − B) = cos²A − sin²B = cos²B − sin²A
- 2sinA cosB = sin(A + B) + sin(A − B)
- 2cosA cosB = cos(A + B) + cos(A − B)
Multiple Angle Formulae
- sin2A = 2sinA cosA = (2 tanA) / (1 + tan²A)
- cos2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1 = (1 − tan²A) / (1 + tan²A)
- sin²A = (1 − cos2A)/2 and cos²A = (1 + cos2A)/2
- tan2A = (2 tanA) / (1 − tan²A)
- tan3A = (3 tanA − tan³A) / (1 − 3 tan²A)
- sin3A = 3sinA − 4sin³A ⇒ sin³A = (3 sinA − sin3A) / 4
- cos3A = 4cos³A − 3cosA ⇒ cos³A = (3 cosA + cos3A) / 4
Trigonometric Ratios of some non-standard angles
- sin15° = cos75° = (√3 − 1) / (2√2)
- cos15° = sin75° = (√3 + 1) / (2√2)
- sin18° = cos72° = (√5 − 1) / 4
- sin54° = cos36° = (√5 + 1) / 4
- sin36° = cos54° = (√10 − 2√5) / 4
- cos18° = sin72° = (√10 + 2√5) / 4
- sin22.5° = ½√(2 − √2)
- cos22.5° = ½√(2 + √2)
- tan18° = 2 − √3
- tan22.5° = √2 − 1
Solved Examples
If sin θ + cosec θ = 2, then sin² θ + cosec² θ is equal to
(A) 1
(B) 4
(C) 2
(D) none of these
Sol. (C). Given sin θ + cosec θ = 2 ; ⇒ (sin θ + cosec θ)² = 4
⇒ sin² θ + cosec² θ + 2 = 4 ; ⇒ sin² θ + cosec² θ = 2.
If (1 + tan θ)(1 + tan ϕ) = 2, then θ + ϕ is equal to
(A) 30°
(B) 45°
(C) 60°
(D) 75°
Sol. Given (1 + tan θ)(1 + tan ϕ) = 2 ; ⇒ 1 + tan θ + tan ϕ + tan θ tan ϕ = 2
⇒ tan θ + tan ϕ = 1 – tan θ tan ϕ ;
------------- = 1 ⇒ tan (θ + ϕ) = 1 ⇒ θ + ϕ = π/4
1 – tan θ tan ϕ
If x = y cos(2π/3) = z cos(4π/3), then xy + yz + zx is equal to
(A) –1
(B) 0
(C) 1
(D) 2
Sol. (B). Given x = y cos(2π/3) = z cos(4π/3); ⇒ x = y (–1/2) = z (–1/2)
⇒ x/1 = y/(–2) = z/(–2) = λ (say); ⇒ x = λ, y = –2λ, z = –2λ.
Hence xy + yz + zx = λ(–2λ) + (–2λ)(–2λ) + (–2λ)λ = 0.
The value of cos(2π/7) + cos(4π/7) + cos(6π/7) + cos(7π/7) is equal to
(A) 1
(B) –1
(C) 1/2
(D) –3/2
Sol. (D). cos(2π/7) + cos(4π/7) + cos(6π/7) + cos(7π/7)
= 1/(2 sin(π/7)) {2 cos(2π/7) sin(π/7) + 2 cos(4π/7) sin(π/7) + 2 cos(6π/7) sin(π/7)} + cos π
= –1/2 – 1 = –3/2
Frequently Asked Questions
Solving trigonometric equations involves a series of systematic steps that make use of identities, algebraic manipulations, and understanding the unit circle. The key steps to solving a basic trigonometric equation are:
- Identify the trigonometric function: Determine whether the equation involves sine, cosine, tangent, or another function.
- Isolate the trigonometric function: If the equation is complex, try to isolate the trigonometric function on one side of the equation.
- Use identities if needed: If the equation is not simple, apply trigonometric identities (like Pythagorean, sum, difference, or double angle formulas) to simplify it.
- Solve for the angle(s): Use inverse trigonometric functions to find the angle(s) that satisfy the equation.
- Consider the domain: Ensure that solutions are within the defined domain for the trigonometric function.
The best approach to solving trigonometric equations depends on the form of the equation, but generally involves:
- Simplification: Break down the equation using trigonometric identities to make the terms simpler.
- Set the trigonometric expression equal to a known value: For example, if you're solving sin
(x) = 0.5, sin(x)=0.5, find the general solutions for x
- Use the unit circle: For most trigonometric equations, the solutions are based on the unit circle, where the angles correspond to specific function values.
- Verify solutions: Always check your solutions by plugging them back into the original equation.
Simplifying trigonometric expressions requires a solid understanding of trigonometric identities and algebraic skills. Start by identifying opportunities to apply common identities, such as the Pythagorean identity
sin2(x) + cos2 (x)=1. Factor or combine terms, and convert between functions (like converting secant to cosine) to simplify the expression as much as possible.
The number of solutions to a trigonometric equation depends on the period of the trigonometric function involved. To determine the number of solutions:
- Identify the function type: For example, sine and cosine have a period of
2π, and tangent has a period ofπ. - Set the equation equal to a specific value: For example, if solving
sin(x) = 0.5, determine how many times the sine curve intersectsy = 0.5within the given domain. - Consider the multiple cycles: If the equation is within an interval larger than one period, multiply the period by the number of complete cycles within that interval.
The general solutions for sine and cosine are derived from their periodic nature. For sine, the general solution is given by:
x = sin⁻¹(y) + 2nπ or x = π - sin⁻¹(y) + 2nπ
For cosine, the general solution is:
x = cos⁻¹(y) + 2nπ or x = -cos⁻¹(y) + 2nπ
Where n is any integer.
To determine periodic solutions, recognize that trigonometric functions repeat after a specific interval (period). The solutions to these equations are all values where the function repeats, so we add integer multiples of the period to the fundamental solution to account for all possible solutions.
The fundamental trigonometric formulas for sine, cosine, and tangent are as follows:
- Sine:
sin(θ) = Opposite / Hypotenuse - Cosine:
cos(θ) = Adjacent / Hypotenuse - Tangent:
tan(θ) = Opposite / Adjacent
These ratios are the building blocks for all other trigonometric identities and function-based formulas.
Graphing trigonometric functions like sine, cosine, and tangent involves understanding their amplitude, period, and phase shifts:
- Amplitude: The maximum displacement from the centerline.
- Period: The length of one complete cycle.
- Phase Shift: Horizontal shift along the x-axis.
By identifying these parameters, you can sketch the graph by plotting key points (like peaks, troughs, and zeros) and ensuring that the graph repeats after one period.