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# Mathematics June 2020 Past Exam Papers and Memos for study revision

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Mathematics Grade 12 June 2020 Exam Papers and Memo

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# Mathematics Grade 12 November 2020 Question Papers and Memos for study revision

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## List Mathematics Grade 12 November 2020 Question Papers and Memos:

Paper 1:

Paper 2:

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Cubic Functions Grade 12 Questions and Answers pdf: In mathematics, a cubic function is a function of the form f(x)=ax^3+bx^2+cx+d where the coefficients a, b, c, and d are real numbers, and the variable x takes real values, and a ≠ 0. In other words, it is both a polynomial function of degree three, and a real function.

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# General Solution Trigonometric Grade 12 pdf

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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.

## 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.

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.

My Courses has a large Questions and Answers repository for the most popular High School and Tertiary Schools subjects. This comes in handy when doing your revision or preparing for exams, tests, research tasks, and assignments.

## 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