TLDR;
This video by Dinesh Sir Live Study provides a comprehensive one-shot lecture on trigonometric functions, covering allied angles, general and principal solutions, and inverse trigonometric functions. It also discusses solution of triangles, including sine rule, cosine rule and projection rule. The lecture includes numerous examples and problem-solving techniques to help students understand and apply the concepts effectively for their mathematics exams.
- Allied angles and their applications in trigonometric functions.
- General and principal solutions of trigonometric equations.
- Sine rule, cosine rule, projection rule and their applications in solving triangles.
- Inverse trigonometric functions, their properties, and problem-solving techniques.
Intro [0:00]
Dinesh Sir welcomes everyone to a one-shot series lecture on trigonometric functions, which is chapter number three of Mathematics One. He mentions that previous chapters like logic, matrices, and pair of straight lines have already been covered. The aim is to complete the chapter in one go, emphasizing the importance of practicing previous chapters as well.
Weightage and Topics [0:55]
The chapter on trigonometric functions carries a weightage of 10 marks with options and 8 marks without options. The topics to be covered include solutions (general and principal), solution of triangles (sine rule, cosine rule, projection rule, Napier's Analogy, and half-angle formula), and inverse trigonometric functions. Questions are expected from every topic, with inverse trigonometric functions being particularly important.
Allied Angles [3:20]
Understanding allied angles is crucial for this chapter. The lecturer uses π and 2π as reference points, avoiding π/2 and 3π/2 to prevent confusion. In the second quadrant, π - theta is used; in the third, π + theta; and in the fourth, 2π - theta. Trigonometric functions that are multiples of π/2 change (sin to cos, tan to cot, sec to cosec), while multiples of π remain the same. The "ASTC" rule (All Silver Tea Cups) is reviewed to remember which trigonometric ratios are positive in each quadrant: all in the first, sin and cosec in the second, tan and cot in the third, and cos and sec in the fourth.
Principal and General Solutions [7:23]
Trigonometric equations have two types of solutions: principal and general. Principal solutions lie between 0 and 2π (excluding 2π), while general solutions are unlimited. Examples of general solutions include sin(nπ) = 0. Nine formulas for general solutions are presented, including those for sin θ = 0, cos θ = 0, tan θ = 0, sin θ = sin α, cos θ = cos α, tan θ = tan α, sin² θ = sin² α, cos² θ = cos² α, and tan² θ = tan² α.
Principal Solution Examples [11:26]
The lecturer demonstrates finding principal solutions with examples. For sin θ = 1/2, the principal solutions are π/6 and 5π/6, both falling within the range of 0 to 2π. For cot x = -√3, the equation is converted to tan x = -1/√3, and the principal solutions are found using allied angles, resulting in 5π/6 and 11π/6.
General Solution Examples [29:08]
The process of finding general solutions is illustrated with examples. For 4 cos² x = 1, the equation is simplified to cos² x = 1/4, then to cos² x = cos²(π/3). The general solution formula θ = nπ ± α is applied, yielding x = nπ ± π/3, where n belongs to the set of integers.
Solving Trigonometric Equations [32:52]
The lecture addresses solving trigonometric equations, emphasizing not to cancel terms to avoid losing solutions. For sin θ = tan θ, the equation is rewritten as sin θ = sin θ / cos θ, leading to sin θ (cos θ - 1) = 0. This results in two equations: sin θ = 0 and cos θ = 1, which are then solved separately to find the general solutions θ = nπ and θ = 2nπ.
Solving a cos x + b sin x = c [37:36]
The method to solve equations of the form a cos x + b sin x = c is explained. The first step involves dividing the entire equation by √(a² + b²). For cos θ + sin θ = 1, the equation is divided by √2, resulting in (1/√2) cos θ + (1/√2) sin θ = 1/√2. The terms are then replaced with trigonometric values, leading to cos(θ - π/4) = 1/√2, which is solved to find the general solution.
Solving sin 5θ + sin θ + sin 3θ = 0 [53:37]
To solve sin 5θ + sin θ + sin 3θ = 0, the combination of small and big angles is used. The formula for sin c + sin d is applied to sin 5θ + sin θ, resulting in 2 sin 3θ cos 2θ + sin 3θ = 0. Factoring out sin 3θ gives sin 3θ (2 cos 2θ + 1) = 0, leading to two equations: sin 3θ = 0 and 2 cos 2θ + 1 = 0. These are solved separately to find the general solutions.
Polar Coordinates and Solution of Triangles [1:05:55]
Polar coordinates are briefly explained, with formulas for converting between polar and Cartesian coordinates. The focus shifts to the solution of triangles, defining terms like semi-perimeter (s = (a + b + c) / 2). The sine rule (a / sin A = b / sin B = c / sin C) and cosine rule (a² = b² + c² - 2bc cos A) are presented, along with projection rules and area of triangle formulas.
Applying Cosine Rule [1:12:51]
An example demonstrates the application of the cosine rule. Starting with a * b cos C - c * cos B, the cosine formulas are substituted, and the expression is simplified to b² - c², proving the given equation.
Applying Projection Rule [1:17:00]
The projection rule is applied to solve a problem. Starting with (A - B cos C) / (B - A cos C), the projection formulas for A and B are substituted, and the expression is simplified to cos B / cos A, proving the given equation.
Applying Sine Rule [1:20:07]
The sine rule is used in conjunction with factorization to solve a problem. Starting with tan((a - b) / 2) = (a - b) / (a + b) * cot(c / 2), the sine rule is applied to express a and b in terms of sines. Factorization formulas are then used to simplify the expression, eventually proving the given equation.
Applying Half Angle Formula [1:31:17]
The half-angle formulas are introduced, and an example is presented to prove sin(A/2) * sin(B/2) * sin(C/2) = (Area of triangle ABC)² / (abcs). The half-angle formulas for sin(A/2), sin(B/2), and sin(C/2) are substituted, and the expression is simplified using Heron's formula for the area of a triangle.
Applying Half Angle Formula with Values [1:32:36]
An example demonstrates the application of half-angle formulas with given values. Given a = 18, b = 24, and c = 30, the semi-perimeter s is calculated. The formula for sin(A/2) is then applied to find its value.
Angles in Arithmetic Progression [1:34:44]
The concept of angles in arithmetic progression (AP) is used to solve a problem. Given that the angles of a triangle ABC are in AP and the ratio of angles B and C is 3:2, the angles are found using the properties of AP and the sine rule.
Inverse Trigonometric Functions: Domain and Range [1:45:59]
The lecture transitions to inverse trigonometric functions, covering their domains and ranges. The ranges for sin⁻¹(x), tan⁻¹(x), and csc⁻¹(x) are between -90° and +90°, while for cos⁻¹(x), cot⁻¹(x), and sec⁻¹(x), they are between 0° and 180°. Special cases and formulas, such as those for tan⁻¹(x) + tan⁻¹(y), are also discussed.
Inverse Trigonometric Functions: Formulas [1:51:11]
Formulas for 2 tan⁻¹(x) are presented, which can be expressed in terms of sin⁻¹, cos⁻¹, or tan⁻¹. Reciprocal identities and formulas for sin⁻¹(x) + cos⁻¹(x), tan⁻¹(x) + cot⁻¹(x), and sec⁻¹(x) + csc⁻¹(x) are also covered.
Principal Value of Inverse Trigonometric Functions [1:52:34]
Examples are provided to find the principal values of inverse trigonometric functions. For sin⁻¹(1/2), the principal value is π/6. For tan⁻¹(-1), the principal value is -π/4. For cos⁻¹(-1/2), the principal value is found using allied angles, resulting in 2π/3.
Solving Expressions with Inverse Trigonometric Functions [2:00:09]
The lecture demonstrates solving expressions involving multiple inverse trigonometric functions. For tan⁻¹(1) + cos⁻¹(1/2) + sin⁻¹(1/2), the individual values are calculated and then summed to find the final answer.
Solving Expressions with Negative Values [2:05:17]
An example is presented to solve an expression with negative values: cos(sin⁻¹(-1/2) + cos⁻¹(-√3/2)). The individual values are calculated using allied angles where necessary, and the expression is simplified to find the final answer.
Applying sin(a+b) Formula [2:11:51]
The formula for sin(a + b) is applied to solve a problem. Starting with sin⁻¹(3/5) + cos⁻¹(12/13), the values of sin x, cos x, sin y, and cos y are found using right-angle triangles. The formula for sin(x + y) is then applied to prove the given equation.
Applying tan(a+b) Formula [2:18:25]
The formula for tan(a + b) is applied to solve a problem. Starting with 2 tan⁻¹(1/3), the expression is rewritten as tan⁻¹(1/3) + tan⁻¹(1/3). The formula for tan(x + y) is then applied to prove the given equation.
Applying tan(a+b) Formula with Values [2:20:57]
The formula for tan(a + b) is applied to solve a problem with values. Starting with tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3), the formula for tan(x + y) is applied to the first two terms, and then again to the resulting expression, eventually proving the given equation.
Outro [2:24:06]
Dinesh Sir concludes the lecture, emphasizing the importance of understanding the concepts and practicing the problems. He encourages students to like and share the video, assuring them that mastering the content will secure 10 solid marks in their exams.