General Solution Trigonometric Grade 12 pdf Download Trigonometry, often deemed as a challenging topic, is a fundamental branch of mathematics that deals with the study of triangles and the relationships between their sides and angles. For high school students, particularly those in Grade 12, mastering trigonometric concepts is crucial not only for academic success but also for laying a solid foundation for advanced studies in mathematics and other STEM fields. Among the key concepts in Grade 12 trigonometry is understanding general solutions, which unlocks a deeper understanding of trigonometric functions and their applications.
What are General Solutions?
In Grade 12 trigonometry, students encounter various types of equations involving trigonometric functions such as sine, cosine, and tangent. These equations often involve angles that satisfy certain conditions, known as constraints. A general solution to a trigonometric equation is a set of all possible angles that satisfy the given equation within a specified range, typically over one or more complete cycles of the trigonometric function.
Understanding the Unit Circle: To understand general solutions in trigonometry, students must first grasp the concept of the unit circle. The unit circle is a circle with a radius of 1 unit, centered at the origin of a Cartesian coordinate system. Angles measured in radians or degrees can be associated with points on the unit circle, allowing for the visualization of trigonometric functions and their values at various angles.
Applying Trigonometric Identities:
Trigonometric identities play a crucial role in finding general solutions to trigonometric equations. These identities, such as the Pythagorean identities and sum and difference identities, allow students to manipulate trigonometric expressions and simplify equations. By applying these identities strategically, students can transform complex trigonometric equations into simpler forms, making it easier to identify general solutions.
Principles of General Solutions:
In Grade 12, students learn several principles for finding general solutions to trigonometric equations. One such principle involves considering the periodic nature of trigonometric functions. Since trigonometric functions repeat their values over regular intervals, general solutions often involve adding multiples of the function’s period to the principal solution.
Another principle is understanding the symmetry of trigonometric functions. By exploiting the symmetrical properties of trigonometric graphs, students can identify additional solutions beyond the principal solution, thereby obtaining a complete set of solutions known as the general solution.
Example:
Consider the trigonometric equation: 2sin(x)=3
. To find the general solution, we first isolate the sine function: sin(x)=32. From the unit circle or trigonometric ratios, we know that sin(π3)=32
. Thus, the principal solution is x=π3.
Since sine has a period of 2π, the general solution is given by x=π3+2πn, where n is an integer representing the number of complete cycles of the sine function.
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 |
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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.
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.
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.