Basic formulae of  Trigonometric Equations and Identities

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Basic formulae of  Trigonometric Equations and Identities

  • A function f(x) is said to be periodic if there exists some T > 0 such that f(x+T) = f(x) for all x in the domain of f(x).
  • In case, the T in the definition of period of f(x) is the smallest positive real number then this ‘T’ is called the period of f(x).
  • Periods of various trigonometric functions are listed below:

1) sin x has period 2π

2) cos x has period 2π

3) tan x has period π

4) sin(ax+b), cos (ax+b), sec(ax+b), cosec (ax+b) all are of period 2π/a

5) tan (ax+b) and cot (ax+b) have π/a as their period

6) |sin (ax+b)|, |cos (ax+b)|, |sec(ax+b)|, |cosec (ax+b)| all are of period π/a

7) |tan (ax+b)| and |cot (ax+b)| have π/2a as their period

  • Sum and Difference Formulae of Trigonometric Ratios

1) sin(a + ß) = sin(a)cos(ß) + cos(a)sin(ß)

2) sin(a – ß) = sin(a)cos(ß) – cos(a)sin(ß)

3) cos(a + ß) = cos(a)cos(ß) – sin(a)sin(ß)

4) cos(a – ß) = cos(a)cos(ß) + sin(a)sin(ß)

5) tan(a + ß) = [tan(a) + tan (ß)]/ [1 – tan(a)tan (ß)]

6)tan(a – ß) = [tan(a) – tan (ß)]/ [1 + tan (a) tan (ß)]

7) tan (π/4 + θ) = (1 + tan θ)/(1 – tan θ)

8) tan (π/4 – θ) = (1 – tan θ)/(1 + tan θ)

9) cot (a + ß) = [cot(a) . cot (ß) – 1]/ [cot (a) +cot (ß)]

10) cot (a – ß) = [cot(a) . cot (ß) + 1]/ [cot (ß) – cot (a)]

  • Double or Triple -Angle Identities

1) sin 2x = 2sin x cos x

2) cos2x = cos2x – sin2x = 1 – 2sin2x = 2cos2x – 1

3) tan 2x = 2 tan x / (1-tan 2x)

4) sin 3x = 3 sin x – 4 sin3x

5) cos3x = 4 cos3x – 3 cosx

6) tan 3x = (3 tan x – tan3x) / (1- 3tan 2x)

  • For angles A, B and C, we have

1) sin (A + B +C) = sinAcosBcosC + cosAsinBcosC + cosAcosBsinC – sinAsinBsinC

2) cos (A + B +C) = cosAcosBcosC- cosAsinBsinC – sinAcosBsinC – sinAsinBcosC

3) tan (A + B +C) = [tan A + tan B + tan C –tan A tan B tan C]/ [1- tan Atan B – tan B tan C –tan A tan C

4) cot (A + B +C) = [cot A cot B cot C – cotA – cot B – cot C]/ [cot A cot B + cot Bcot C +  cot A cotC–1]

  • List of some other trigonometric formulas:

1) 2sinAcosB = sin(A + B) + sin (A – B)

2) 2cosAsinB = sin(A + B) – sin (A – B)

3) 2cosAcosB = cos(A + B) + cos(A – B)

4) 2sinAsinB = cos(A – B) – cos (A + B)

5) sin A + sin  B = 2 sin [(A+B)/2] cos [(A-B)/2]

6) sin A – sin  B = 2 sin [(A-B)/2] cos [(A+B)/2]

7) cosA + cos  B = 2 cos [(A+B)/2] cos [(A-B)/2]

8) cosA – cos  B = 2 sin [(A+B)/2] sin [(B-A)/2]

9) tanA ± tanB = sin (A ± B)/ cos A cos B

10)cot A ± cot B = sin (B ± A)/ sin A sin B

  • Method of solving a trigonometric equation:

1) If possible, reduce the equation in terms of any one variable, preferably x. Then solve the equation as you used to in case of a single variable.

2) Try to derive the linear/algebraic simultaneous equations from the given trigonometric equations and solve them as algebraic simultaneous equations.

3) At times, you might be required to make certain substitutions. It would be beneficial when the system has only two trigonometric functions.

  • Some results which are useful for solving trigonometric equations:

1) sin θ = sina and cosθ = cosa ⇒ θ = 2nπ + a

2) sin θ = 0 ⇒ θ = nπ

3) cosθ = 0 ⇒ θ = (2n + 1)π/2

4) tan θ = 0 ⇒ θ = nπ

5) sinθ = sina⇒ θ = nπ + (-1)na where a ∈ [–π/2, π/2]

6) cosθ= cos a ⇒ θ = 2nπ ± a, where a ∈[0,π]

7) tanθ = tana⇒ θ = nπ+ a, where a ∈[–π/2, π/2]

8) sinθ = 1 ⇒ θ= (4n + 1)π/2

9) sin θ = -1 ⇒ θ = (4n – 1) π /2

10) sin θ = -1 ⇒ θ = (2n +1) π /2

11) |sinθ| = 1⇒ θ =2nπ

12) cosθ = 1 ⇒ θ =(2n + 1)

13) |cosθ| = 1⇒ θ =nπ

 

Solved Examples on Trigonometric Equations and Identities

Illustration 1: If cos (a + b) = 4/5, sin (a-b) = 5/13 and a and b lie between 0 to π/4, find tan 2a.

Solution: It is given that cos (a + b) = 4/5, sin (a-b) = 5/13

It follows that sin (a + b) = 3/5, cos (a-b) = 12/13

Hence, tan (a+b) = 3/4 and tan (a-b) = 5/12

Hence, this implies

tan [(a+b) + a-b] =[tan (a+b) + tan (a-b)]/ [1 + tan(a+b)tan(a-b)]

= (3/4 + 5/12)/ (1 – 3/4. 5/12)

Hence, tan 2a = 14/12. 48/33 = 56/33.

 

Illustration 2: If A + B + C = 1800, prove that

sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C.

Solution:We have

sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C

So consider LHS = sin (B + C – A) + sin (C + A – B) + sin (A + B – C)

= sin (π– A – A) + sin (π – B – B) + sin (π– C – C)(since A + B + C = π)

= sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C

Illustration 3: Solve the equation 5 sinθ – 2 cos2θ – 1 = 0

Solution: Given, 5 sinθ – 2 cos2θ – 1 = 0

or,5 sinθ – 2 (1 – sin2θ) – 1 = 0

or,2 sin2 θ + 5 sin θ – 3 = 0

or,(sin θ + 3) (2 sin θ – 1) = 0

∴sin θ = 3 or, sin θ = 1/2

if sin θ = 3

This is not possible as range of sine is [–1, 1]

If sin θ = 1/2

or, sin θ = sin π/6

⇒ θ = n π + (–1)n π/6

where n = 0, ±1, ±2 ……