General Solution Trigonometric Grade 12 pdf: This page contains a guide for Mathematics Grade 12 Students with the following Trigonometric notes: Trigonometric Functions, 3D Trigonometry, Identities Problems with Solutions, and Challenging Exam Practices.
About Mathematics Grade 12
About Grade 12 Mathematics: Mathematics is a language that makes use of symbols and notations for describing numerical, geometric and graphical relationships. It is a human activity that involves observing, representing and investigating patterns and qualitative relationships in physical and social phenomena and between mathematical objects themselves. It helps to developmental processes that enhance logical and critical thinking, accuracy and problem solving that will contribute in decision-making. Mathematical problem solving enables us to understand the world (physical, social and economic) around us, and, most of all, to teach us to think creatively.
View all #Mathematics-Grade 12 Study Resources
We have compiled great resources for Mathematics Grade 12 students in one place. Find all Question Papers, Notes, Previous Tests, Annual Teaching Plans, and CAPS Documents.
Topics covered in Mathematics Grade 12
Below are the topics covered in Mathematics Grade 12:
- Patterns, Sequences and Series
- Functions and Interverse Functions
- Exponential and Logarithmic Functions
- Finance, Growth and Decay
- Trigonometry: Compound and Double Angle Identities
- Trigonometry: Problem Solving in Two and Three Dimentions
- Polynomials
- Differential Calculus
- Analytical Geometry
- Euclidean Geometry
- Statistics
- Counting Principles and Probability
Concepts and skills
- Simplify trigonometric equations.
- Find the reference angle.
- Use reduction formulae to find other angles within each quadrant.
- Find the general solution of a given trigonometric equation.
Prior knowledge:
- Trigonometric ratios.
- Trigonometric identities.
- Solve problems in the Cartesian plane.
- Co-functions.
- Factorisation.
- Revise the following trigonometric ratios for right-angled triangles, which you learnt in Grade 10.
Will start with the basic equation, sin x = 0. The principal solution for this case will be x = 0, π, 2π as these values satisfy the given equation lying in the interval [0, 2π]. But, we know that if sin x = 0, then x = 0, π, 2π, π, -2π, -6π, etc. are solutions of the given equation. Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.
Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, n∈I, as cos x has a value equal to 0 at π/2, 3π/2, 5π/2, -7π/2, -11π/2 etc. Below here is the table defining the general solutions of the given trigonometric functions involved equations.
Solutions Trigonometric Equations
| Equations | Solutions |
| sin x = 0 | x = nπ |
| cos x = 0 | x = (nπ + π/2) |
| tan x = 0 | x = nπ |
| sin x = 1 | x = (2nπ + π/2) = (4n+1)π/2 |
| cos x = 1 | x = 2nπ |
| sin x = sin θ | x = nπ + (-1)nθ, where θ ∈ [-π/2, π/2] |
| cos x = cos θ | x = 2nπ ± θ, where θ ∈ (0, π] |
| tan x = tan θ | x = nπ + θ, where θ ∈ (-π/2 , π/2] |
| sin2 x = sin2 θ | x = nπ ± θ |
| cos2 x = cos2 θ | x = nπ ± θ |
| tan2 x = tan2 θ | x = nπ ± θ |
Watch: Video
How to solve general trigonometric equations formula and find solutions
The periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. If we do not restrict the solution, then we need to determine the general solution to the equation. We know that the sine and cosine functions have a period of 360360° and the tangent function has a period of 180180°.
Method for finding the solution:
- Simplify the equation using algebraic methods and trigonometric identities.
- Determine the reference angle (use a positive value).
- Use the CAST diagram to determine where the function is positive or negative (depending on the given equation/information).
- Restricted values: find the angles that lie within a specified interval by adding/subtracting multiples of the appropriate period.
- General solution: find the angles in the interval [0°;360°][0°;360°] that satisfy the equation and add multiples of the period to each answer.
- Check answers using a calculator.
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Questions and Answers
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Identities Problems with Solutions for exam Practice
1. (1 – sin A)/(1 + sin A) = (sec A – tan A)2
Solution:
L.H.S = (1 – sin A)/(1 + sin A)
= (1 – sin A)2/(1 – sin A) (1 + sin A),[Multiply both numerator and denominator by (1 – sin A)
= (1 – sin A)2/(1 – sin2 A)
= (1 – sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 – sin2 θ]
= {(1 – sin A)/cos A}2
= (1/cos A – sin A/cos A)2
= (sec A – tan A)2 = R.H.S. Proved.
2. Prove that, √{(sec θ – 1)/(sec θ + 1)} = cosec θ – cot θ.
Solution:
L.H.S.= √{(sec θ – 1)/(sec θ + 1)}
= √[{(sec θ – 1) (sec θ – 1)}/{(sec θ + 1) (sec θ – 1)}]; [multiplying numerator and denominator by (sec θ – l) under radical sign]
= √{(sec θ – 1)2/(sec2 θ – 1)}
=√{(sec θ -1)2/tan2 θ}; [since, sec2 θ = 1 + tan2 θ ⇒ sec2 θ – 1 = tan2 θ]
= (sec θ – 1)/tan θ
= (sec θ/tan θ) – (1/tan θ)
= {(1/cos θ)/(sin θ/cos θ)} – cot θ
= {(1/cos θ) × (cos θ/sin θ)} – cot θ
= (1/sin θ) – cot θ
= cosec θ – cot θ = R.H.S. Proved.
3. tan4 θ + tan2 θ = sec4 θ – sec2 θ
Solution:
L.H.S = tan4 θ + tan2 θ
= tan2 θ (tan2 θ + 1)
= (sec2 θ – 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]
= (sec2 θ – 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]
= sec4 θ – sec2 θ = R.H.S. Proved.
More problems on trigonometric identities are shown where one side of the identity ends up with the other side.
4. . cos θ/(1 – tan θ) + sin θ/(1 – cot θ) = sin θ + cos θ
Solution:
L.H.S = cos θ/(1 – tan θ) + sin θ/(1 – cot θ)
= cos θ/{1 – (sin θ/cos θ)} + sin θ/{1 – (cos θ/sin θ)}
= cos θ/{(cos θ – sin θ)/cos θ} + sin θ/{(sin θ – cos θ/sin θ)}
= cos2 θ/(cos θ – sin θ) + sin2 θ/(cos θ – sin θ)
= (cos2 θ – sin2 θ)/(cos θ – sin θ)
= [(cos θ + sin θ)(cos θ – sin θ)]/(cos θ – sin θ)
= (cos θ + sin θ) = R.H.S. Proved.
5. Show that, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A)
Solution:
We have,
1/(csc A – cot A) + 1/(csc A + cot A)
= (csc A + cot A + csc A – cot A)/(csc2 A – cot2 A)
= (2 csc A)/1; [since, csc2 A = 1 + cot2 A ⇒ csc2A – cot2 A = 1]
= 2/sin A; [since, csc A = 1/sin A]
Therefore,
1/(csc A – cot A) + 1/(csc A + cot A) = 2/sin A
⇒ 1/(csc A – cot A) + 1/(csc A + cot A) = 1/sin A + 1/sin A
Therefore, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A) Proved.
6. (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ)/cos θ
Solution:
L.H.S = (tan θ + sec θ – 1)/(tan θ – sec θ + 1)
= [(tan θ + sec θ) – (sec2 θ – tan2 θ)]/(tan θ – sec θ + 1), [Since, sec2 θ – tan2 θ = 1]
= {(tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ)}/(tan θ – sec θ + 1)
= {(tan θ + sec θ) (1 – sec θ + tan θ)}/(tan θ – sec θ + 1)
= {(tan θ + sec θ) (tan θ – sec θ + 1)}/(tan θ – sec θ + 1)
= tan θ + sec θ
= (sin θ/cos θ) + (1/cos θ)
= (sin θ + 1)/cos θ
= (1 + sin θ)/cos θ = R.H.S. Proved.
Sources:
https://www.math-only-math.com/problems-on-trigonometric-identities.html
https://byjus.com/maths/trigonometric-equations/
