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# Trigonometric Functions and Solutions Paper

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Trigonometry

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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
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 nI.
Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, nI, 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
How to solve general trigonometric equations
formula and find solutions

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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.
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)}

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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 ...
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Anonymous
Awesome! Perfect study aid.

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