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

Basic Trigonometric Identities

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

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:

1. sinC + sinD = 2 sin (C + D/2) cos (C - D/2)
2. sinC - sinD = 2 cos (C + D/2) sin (C - D/2)
3. cosC + cosD = 2 cos (C + D/2) cos (C - D/2)
4. cosC - cosD = - 2 sin (C + D/2) sin (C - D/2)

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

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)

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.
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• 
• cos 3A = 4 cos³A - 3 cosA
• sin 3A = 3 sinA - 4 sin³A
•  

Important Trigonometric Ratios

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 exists 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