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Chapter 9: Solving Trigonometric Equations
Summary
The overall substance of this chapter is to solve equations which contain trig functions, for all values of an angle in a given
range. The range can be expressed as degrees or radians, which dictates whether the answer should be expressed as degrees
or radians.
Steps to solving trig equations for all angle values:
Step #1: Isolate the Function
Methods to isolating the function: (Decimals are okay here!)
ο‚· Simple Solving: Add/Subtract or Multiple/Divide or take a square root to get the function alone
Example:
2 sin πœƒ + 3 = 4
(subtract 3 from both sides)
2 sin πœƒ = 1
(divide both sides by 2)
1
sin πœƒ =
Isolated 
2
ο‚·
Solve by GCF factoring: Is the term a binomial with a common factor? Set to zero, factor completely, set each factor equal
to zero; finish by simple solving
Example:
2 cos 2 πœƒ βˆ’ cos πœƒ = 0
cos πœƒ (2 cos πœƒ βˆ’ 1) = 0
cos πœƒ = 0
2 cos πœƒ βˆ’ 1 = 0
1
cos πœƒ = 0
cos πœƒ =
Isolated 
2
ο‚·
Solve by Case I or Case II trinomial factoring: Is the term a trinomial? Set to zero, factor completely, set each factor equal to
zero; finish by simple solving.
Example:
tan2 πœƒ βˆ’ tan πœƒ βˆ’ 12 = 0
(tan πœƒ βˆ’ 4)(tan πœƒ + 3) = 0
tan πœƒ βˆ’ 4 = 0
tan πœƒ + 3 = 0
tan πœƒ = 4 tan πœƒ = βˆ’3
Isolated 
ο‚·
Solve by quadratic formula: Is it a trinomial? Quadratic formula is always an option as long as the equation is a quadratic
equation (π‘Žπ‘₯ 2 + 𝑏π‘₯ = 𝑐 = 0). Set equal to zero, plug in and evaluate carefully, split the ±, get values (decimal is possible)
for both terms.
Example:
sin2 πœƒ + 3 sin πœƒ βˆ’ 8 = 0
sin πœƒ =
sin πœƒ =
sin πœƒ =
sin πœƒ =
βˆ’π‘±βˆšπ‘ 2 βˆ’4π‘Žπ‘
2π‘Ž
βˆ’3±βˆš32 βˆ’4(1)(βˆ’8)
2(1)
βˆ’3±βˆš41
2
βˆ’3+√41
2
sin πœƒ = 1.702
ο‚·
sin πœƒ =
βˆ’3βˆ’βˆš41
2
sin πœƒ = βˆ’4.702
Isolated 
Solve using reciprocal functions: Does the expression contain more than one trig monomial? Find a way to rewrite one
(probably will involve cross multiplying). Solve using one of the above methods.
Example:
2 sin πœƒ = csc πœƒ
2 sin πœƒ
1
=
1
sin πœƒ
2 sin2 πœƒ = 1
sin2 πœƒ =
1
2
sin πœƒ = ±βˆš
sin πœƒ = √
1
2
1
2
sin πœƒ = 0.25
sin πœƒ = βˆ’βˆš
1
2
sin πœƒ = βˆ’0.25
Isolated 
ο‚·
Solve using Pythagorean trig identities: Does the expression contain more than one trig function mixed in a binomial or
trinomial? Is one of the terms squared? Use Pythagorean identities to change the squared term. Solve using one of the
methods above.
Memorize three Pythagorean Identities
(1) 𝑠𝑖𝑛2 πœƒ + π‘π‘œπ‘  2 πœƒ = 1
(2) π‘‘π‘Žπ‘›2 πœƒ + 1 = 𝑠𝑒𝑐 2 πœƒ
(3) π‘π‘œπ‘‘ 2 πœƒ + 1 = 𝑐𝑠𝑐 2 πœƒ
2
Example:
cos πœƒ + sin πœƒ βˆ’ 1 = 0
1 βˆ’ sin2 πœƒ + sin πœƒ βˆ’ 1 = 0
βˆ’ sin2 πœƒ + sin πœƒ = 0
sin2 πœƒ βˆ’ sin πœƒ = 0
sin πœƒ (sin πœƒ βˆ’ 1) = 0
sin πœƒ = 0
sin πœƒ βˆ’ 1 = 0
sin πœƒ = 0
sin πœƒ = 1
Isolated 
ο‚·
Solve using Double Angle formulas: Does one of the terms contain a double angle? Replace it using one of the double angle
formulas (provided on the reference sheet). Solve using one of the methods above.
Example:
sin 2πœƒ + cos πœƒ = 0
2 sin πœƒ cos πœƒ + cos πœƒ = 0
cos πœƒ (2 sin πœƒ + 1) = 0
cos πœƒ = 0
2 sin πœƒ + 1 = 0
1
cos πœƒ = 0
sin πœƒ = βˆ’
Isolated 
2
Step #2: Find the reference angle
ο‚·
Check First: If the function is sine or cosine, and the result is βˆ’1, 0 or 1,
then your answer is a quadrantial angle (0°, 90°, 180°, 270°, 360°).
Find it, and jump right to step #5.
To determine the correct quadrantial angle, remember the unit circle.
Cosine represents x-values, and Sine represents y-values
Example: sin πœƒ = βˆ’1
Sine represents the y-value.
Picture the value -1 on the y-axis.
What angle is in that direction? 270°
ο‚·
To find a reference angle, look at the POSITIVE value of the isolated function from step #1. Use your calculator’s inverse
function. Round to the nearest tenth unless otherwise instructed.
3
Example:
sin πœƒ = βˆ’
4
3
Calculator input: sinβˆ’1 ( )
Reference angle = 48.6°
4
If you have a reciprocal function, find the reciprocal number first (flip the term), then ask your calculator
4
Example:
cot πœƒ =
tan πœƒ =
5
5
4
5
Calculator input: tanβˆ’1 ( )
Reference angle = 51.3°
4
Step #3 & #4: Choose quadrants & Find the full angles
ο‚·
ο‚·
Step #3 is when you care about whether the isolated function in step #1 is positive or negative. Remember ASTC and
choose TWO quadrants that fit the question. Draw a simple coordinate plane, and place an angle in the correct quadrant.
Squeeze the reference angle from step #2 to the x-axis (never the y!)
In Step #4, find the full measure of each angle drawn, starting from 0°. Think about how far away your angle is from the xaxis (never the y!) based on the reference angle squeezed in there.
1
3
Example
sin πœƒ =
sin πœƒ = βˆ’
2
Ref angle = 30°
(ASTC says sine is positive in I and II
4
Ref angle = 48.6°
sine is negative in III and IV)
Step #5: Check ALL answers in your calculator
Plugging in very carefully, check that each solution (not just one solution) checks properly. Since each solution is arrived at
individually (different reference angles, different quadrants) each one must be checked.
Example:
tan2 πœƒ + sec πœƒ = βˆ’1
Would result in an answer of 180°
(tan(180))2 + (1/ cos(180))
To test 180°, input:
Does it equal βˆ’1 like the original question asked? οƒΌ
The answer is correct and should be included in the solution set for the question.
Repeat the check process for EACH answer from step #4.
VERY RARELY will there actually be an extraneous solution (a correctly processed answer that doesn’t fit the original
question). If a check doesn’t work, odds are very good that you made a mistake along the way. Go back and look for your error.
Your final answer should be a solution set containing all answer choices that check out.
If the question was asked with a degree range (Example: Find all values of πœƒ such that 0° ≀ πœƒ < 360°), you’re done.
Sample solution expressed in degrees:
πœƒ = {30°, 150°, 200°, 340°}
If the question was asked with a radian range (Example: Fins all values of πœƒ such that 0 ≀ πœƒ < 2πœ‹), convert each answer.
πœ‹
Recall how to convert: radians = π‘‘π‘’π‘”π‘Ÿπ‘’π‘’ β‹…
180
Same sample solution expressed in radians:
πœ‹ 5πœ‹ 10πœ‹ 17πœ‹
πœƒ={ ,
6
6
,
9
,
9
}