Important Formulas & Examples: Trigonometric Functions - 1 - Mathematics (Maths) Class 11 - Commerce

In mathematics, trigonometry is one of the most important topics to learn. Trigonometry is the study of triangles, where "Trigon" means triangle and "metry" means measurement. Below is a list of formulas based on the right triangle that can be used for studying trigonometry.

Trigonometric Ratios

The three main functions in trigonometry are Sine, Cosine and Tangent. They are just the length of one side divided by another

Consider a right-angled triangle with an angle θ, a hypotenuse, a side adjacent to angle θ, and a side opposite the angle to angle θ. For a given angle θ each ratio stays the same no matter how big or small the triangle is

• Sine Function: sin(θ) = Opposite Side / Hypotenuse
• Cosine Function: cos(θ) = Adjacent Side / Hypotenuse
• Tangent Function: tan(θ) = Opposite Side / Adjacent Side

When we divide Sine by Cosine we get:

• Cosecant Function: cosec(θ) = Hypotenuse / Opposite

• Secant Function: sec(θ) = Hypotenuse / Adjacent

• Cotangent Function: cot(θ) = Adjacent / Opposite

Trigonometric Ratios Table

The trigonometric table is simply a collection of the values of trigonometric ratios for various angles including 0°, 30°, 45°, 60°, 90°, sometimes with other angles like 180°, 270°, and 360° included, in a tabular format.
Here is the trigonometry table for standard angles along with some non-standard angles:

Basic Trigonometric Identities

Tangent and Cotangent Identities

• tan(θ) = sin(θ) / cos(θ)
• cot(θ) = cos(θ) / sin(θ)

Reciprocal Identities

• sin(θ)= 1/cosec(θ)
• cosec(θ) = 1/sin(θ)
• cos(θ) = 1/sec(θ)
• sec(θ) = 1/cos(θ)
• tan(θ) = 1/cot(θ)
• cot(θ) = 1/tan(θ)

Pythagorean Identities

• sin²(θ) + cos²(θ) = 1 ; (-1 ≤ sin θ ≤ 1 ; -1 ≤ cos θ ≤ 1 , θ ∈ R)
• 1 + tan²(θ) = sec²(θ) ; ( |sec θ| ≥ 1, θ ∈ R)
• 1 + cot²(θ) = cosec²(θ)  ; (|cosec θ| ≥ 1 , θ ∈ R)

Even and Odd Angle Formulas

• sin(-θ) = -sin(θ)
• cos(-θ) = cos(θ)
• tan(-θ) = -tan(θ)
• cot(-θ) = -cot(θ)
• sec(-θ) = sec(θ)
• cosec(-θ) = -cosec(θ)

Co-function Formulas

• sin(90°-θ) = cos(θ)
• cos(90°-θ) = sin(θ)
• tan(90°-θ) = cot(θ)
• cot(90°-θ) = tan(θ)
• sec(90°-θ) = cosec(θ)
• cosec(90°-θ) = sec(θ)

Important Trigonometric Ratios

• sin nπ = 0 ; cos nπ = (-1)ⁿ ; tan nπ = 0 where n ∈ I
• 

Trigonometric Functions of Allied Angles

If θ is any angle , then - θ, 90 ±θ , 180 ± θ, 270 ± θ , 360 ± θ etc. are called ALLIED ANGLES.

 Example 1. Express 1·2 radians in degree measure.

1·2 radians =  

Example 2. Calculate sin α if cos α = - 9/11 and α ∈ (π, 3π/2).

For any angle a belonging to the indicated interval sin a is negative, and therefore,

Example 3. Calculate tan α if cos α = √5/5 and α ∈ (π, 3π/2)

For any angle a belonging to the indicated interval tanα is positive and cos a is negative, and therefore

Trigonometric Functions of Sum Or Difference of Two Angles

• sin (A ± B) = sinA cosB ± cosA sinB
• cos (A ± B) = cosA cosB m sinA sinB
• sin²A - sin²B = cos²B - cos²A = sin (A+B) . sin (A- B)
• cos²A - sin²B = cos²B - sin²A = cos (A+B) . cos (A - B)
• 

 Factorisation of the Sum Or Difference of Two sines Or cosines:

• sinC + sinD = 2 sin (C + D/2) cos (C - D/2)
• sinC - sinD = 2 cos (C + D/2) sin (C - D/2)
• cosC + cosD = 2 cos (C + D/2) cos (C - D/2)
• cosC - cosD = - 2 sin (C + D/2) sin (C - D/2)

Transformation of Products Into Sum Or Difference Of sines & cosines:

• 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 4. Suppose x and y are real numbers such that tan x + tan y = 42 and cot x + cot y = 49. Find the value of tan(x + y).

tan x + tan y = 42 and cot x + cot y = 49

now, cot x + cot y = 49 

Example 5. If x sinθ = y sin(θ + 2π/3) = z sin (θ + 4π/3) then:
(a) x + y + z = 0 
(b) xy + yz + zx = 0 
(c) xyz + x + y + z = 1 
(d) none

Correct Answer is Option (b)

= xy + yz + zx = 0

Example 6. Find θ satisfying the equation, tan 15° . tan 25° . tan 35° = tan 0, where θ ∈ (0, 15°). 

LHS = tan 15° . tan (30° - 5°) . tan (30° + 5°)
let t = tan 30° and m = tan 5°

 Example 7. If tan A & tan B are the roots of the quadratic equation, a x² + b x + c = 0 then evaluate a sin² (A + B) + b sin (A + B) . cos (A + B) + c cos² (A + B).

Now E = cos² (A + B) [a tan² (A + B) + b tan (A + B) + c]

Example 8. Show that cos²A + cos²(A + B) + 2 cosA cos(180° + B) · cos(360° + A + B) is independent of A. Hence find its value when B = 810°.

cos²A + cos²(A + B) - [2 cosA · cosB · cos (A + B)]
= cos²A + cos²(A + B) - [ {cos(A + B) + cos(A - B) } cos (A + B) ]
= cos²A + cos²(A + B) - cos²(A + B) - (cos²A - sin²B)
= sin²B which is independent of A
now,  sin²(810°) = sin²(720° + 90°) = sin²90° = 1

Multiple Angles & Sub-Multiple Angles

• sin 2A = 2 sinA cosA ; sin θ = 2 Sin(θ/2)Cos(θ/2)
• cos 2A = cos²A - sin²A = 2cos²A - 1 = 1 - 2 sin²A ;
  cos θ= cos² θ/2 - sin² θ/2 = 2cos² θ/2 - 1 = 1 - 2sin² θ/2.
  2 cos²A = 1 + cos 2A , 2sin²A = 1 - cos 2A
  2 cos² θ/2 = 1 + cos θ, 2 sin² θ/2 = 1 - cos θ.
•  
• 
• cos 3A = 4 cos³A - 3 cosA
• sin 3A = 3 sinA - 4 sin³A
•  

Conditional Identities

If A+B+C = π

• tanA + tanB + tanC = tanA tanB tanC
• 
• sin2A + sin2B + sin2C = 4 sinA sinB sinC
• sinA + sinB + sinC = 4 cos A/2 Cos B/2 Cos C/2
• cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C
• cos A + cos B + cos C = 1 + 4 sin A/2 sin B/2 sin C/2

Example 10. If A + B + C = π, prove that,

Solution:

Example 11. If A + B + C = θ and cotθ = cot A + cot B + cot C, show that , sin(A-θ). sin (B-θ).sin (C-θ)= sin³ θ.

Solution:

Given cot θ = cot A + cot B + cot C or cot θ - cot A = cot B + cot C

 ....(1)

....(2)

....(3)

Multiplying (1) , (2) and (3) we get the result

Example 12. Find whether a triangle ABC can exist with the tangents of its interior angle satisfying, tan A = x, tan B = x + 1, and tan C = 1 - x for some real value of x. Justify your assertion with adequate reasoning.

Solution:

In a triangle ∑ tan A = π tan A (to be proved)
x + x + 1 + 1 – x = x(1 + x)(1 – x)
2 + x = x – x³; x³ = –2, x = -2^1/3
Hence tanA = x < 0 and tanB = x + 1 = 1 – 2^1/3 < 0
Hence A and B both are obtuse. Which is not possible in a triangle. Hence no such triangle can exist.

Maximum & Minimum values of Trigonometric Functions

• Min. value of a² tan² θ + b² cot²θ = 2ab
• Max and Min. value of acosθ + bsinθ are  
• If f(θ) = acos(α + q) + bcos(β + q) where a, b, α and β are known quantities then
  
•    (constant) then the maximum values of the expression cosα cosβ, cosα+ cosβ, sinα + sinβ and sina sinb occurs when α = β = σ/2
• If A, B, C are the anlges of a triangle then maximum value of sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 60º
• In case a quadratic in sinq or cosq is given then the maximum or minimum values can be interpreted by making a perfect square

Example 13. Find the greatest value of c such that system of equations x² + y² = 25; x + y = c has a real solution.

Solution:

Put x = 5 cosθ and y = 5 sinθ

5(cosθ+ sinθ) = c; but (cosθ + sinθ)max = √2 and (cosθ + sinθ)min = – √2

hence, c_max = 5√2

Example 14. Find the minimum and maximum value of f (x, y) = 7x² + 4xy + 3y² subjected to x² + y² = 1.

Solution:

Let x = cosθ and y = sinθ
y = f (θ) = 7 cos2θ + 4 sin θcosθ + 3 sin2θ = 3 + 2 sin 2q + 2(1 + cos 2θ)
= 5 + 2(sin 2θ + cos 2θ) but -√2 ≤ (sin 2θ + cos 2θ) ≤ √2
y_max = 5 + 2√2 and y_min = 5 – 2√2