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General Solution Trigonometric Grade 12 pdf

General Solution Trigonometric Grade 12 pdf Typical Resolution PDF for trigonometry in grade 12: Trigonometric Functions, 3D Trigonometry, Identities Problems with Solutions, and Challenging Exam Practices are all covered in this page’s resource for mathematics students in grade 12.

Mathematics Grade 12

Mathematics for Grade 12: Mathematics is a language that uses symbols and notations to describe relationships in numbers, geometry, and graphics. It is a human activity that involves noticing, expressing, and analyzing qualitative connections between mathematical objects as well as between physical and social occurrences. It aids in the growth of cognitive processes that improve accuracy, logical and critical reasoning, and problem-solving abilities that support decision-making. Solving mathematical puzzles helps us to comprehend the physical, social, and economic world around us and, most importantly, it develops our capacity for original thought.

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:

1. Simplify the equation using algebraic methods and trigonometric identities.
2. Determine the reference angle (use a positive value).
3. Use the CAST diagram to determine where the function is positive or negative (depending on the given equation/information).
4. Restricted values: find the angles that lie within a specified interval by adding/subtracting multiples of the appropriate period.
5. 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.
6. Check answers using a calculator.

Download General Solution Trigonometric Grade 12 pdf

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.

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.

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