30.1 Trigonometric Identities Concept Overview
TRIGONOMETRIC IDENTITIES | CONCEPT OVERVIEW The TOPIC of TRIGONOMETRIC IDENTITIES can be referenced on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: An IDENTITY is an expression or formula that is always true. Identities are used to solve equations as they can be multiplied on both sides of the equation. A TRIGONOMETRIC IDENTITY is an equation involving trigonometric functions that can be solved by any angle. Trigonometric identities have less to do with evaluating functions at specific angles and more to do with relationships between functions. During the exam, all of the TRIGONOMETRIC IDENTITIES you will ever need can be found in the NCEES SUPPLIED REFERENCE HANDBOOK. There are some general steps that can be followed when solving equations that use trigonometric identities: STEP 1: • Look for obvious simplifications. This should be your first step in solving any equation. Made with
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• If you can easily see one of the identities right away, go ahead and start substituting the identity into the equation to start simplifying terms. • If you have an expression with many different trig functions inside, try to think if it can be simplified in any other method obeying the laws of algebraic expressions. For example, can you use the conjugate or another identity to simplify terms. STEP 2: • Convert as many terms as you can to sine and cosine. You can do this using reciprocal identities, tangent identities, and cotangent identities. • Sine and cosine terms are easier to work with, and have identities that are easier to identify when being applied. • See if you can now combine like terms, eliminate fractions, and simplify the expressions on each side of the equation. • Try to identify what terms or expressions can be factored out and replaced with trigonometric identities.
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STEP 3: • Go through all of the trigonometric identities, and see if you can apply any of them. Pay attention to how terms are multiplied, what power terms are raised to, and the +/- of each term to figure out the best trigonometric identity to use. • The most commonly use trigonometric identities on the exam are the PYTHOGREAN IDENTITIES. STEP 4: • The final step is to make sure your answer is one of the answer choices. It is common to do everything correctly, and get your answer in a different form than that of one of the answer choices. • Don’t worry! You have the right answer, you may just need to apply another identity, or re-arrange your answer. • If you get to this point, and still can’t get an answer choice, convert everything to sine and cosine terms again. RECIPROCAL IDENTITIES: The GENERAL FORMULAS highlighting the RECIPROCAL IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with
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The can be stated as:
sin 𝜃 =
1 csc 𝜃
cos 𝜃 =
1 sec 𝜃
tan 𝜃 =
1 cot 𝜃
csc 𝜃 =
1 sin 𝜃
sec 𝜃 =
1 cos 𝜃
cot 𝜃 =
1 tan 𝜃
RATIO IDENTITIES: The FORMULAS for the RATIO IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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They can be stated as:
tan 𝜃 =
sin 𝜃 cos 𝜃
cot 𝜃 =
cos 𝜃 sin 𝜃
PYTHAGOREAN IDENTITIES: The FORMULAS for the PYTHAGOREAN IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. They are stated as: sin2 𝜃 + cos 2 𝜃 = 1 tan2 𝜃 + 1 = sec 2 𝜃 cot 2 𝜃 + 1 = csc 2 𝜃
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Perform the addition and simplification of the following trigonometric expression: sin 𝜃 cos 𝜃 + 1 + cos 𝜃 sin 𝜃 A. tan 𝜃 B. csc 𝜃 C. cot 𝜃 D. sin 𝜃
SOLUTION: We are given the expression originally as: sin 𝜃 cos 𝜃 + 1 + cos 𝜃 sin 𝜃 We now need to look for ways to expand, rearrange, combine…whatever we can do to simplify it the best that we can.
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Establishing the common denominator, we are able to pull these two terms together such that: sin 𝜃 cos 𝜃 (sin 𝜃)(sin 𝜃) + (cos 𝜃)(1 + cos 𝜃) + = 1 + cos 𝜃 sin 𝜃 (1 + cos 𝜃)(sin 𝜃) The sine and cosine terms are now beginning to reflect a trigonometric identity, which is what we want in route to simplifying. Multiplying out the factored terms in the numerator, we get: sin2 𝜃 + cos 2 𝜃 + cos 𝜃 (1 + cos 𝜃)(sin 𝜃) The most commonly used trigonometric identities that we will encounter on the exam are those of the PYTHOGREAN IDENTITIES. The FORMULAS for the PYTHAGOREAN IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The PYTHAGOREAN IDENTITIES state that: sin2 𝜃 + cos 2 𝜃 = 1 tan2 𝜃 + 1 = sec 2 𝜃
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cot 2 𝜃 + 1 = csc 2 𝜃 Heading back to the expression we have at this point, we can simplify some terms into a Pythagorean identity, such that: sin2 𝜃 + cos 2 𝜃 + cos 𝜃 (1 + cos 𝜃) = (1 + cos 𝜃)(sin 𝜃) (1 + cos 𝜃)(sin 𝜃) With both (1 + cos 𝜃) in the numerator and the denominator, let’s divide it out, leaving us with: (1 + cos 𝜃) 1 = (1 + cos 𝜃)(sin 𝜃) sin 𝜃 Using the reciprocal identity, our final answer is: 1 = csc 𝜃 sin 𝜃 The correct answer choice is B. 𝒄𝒔𝒄 𝜽
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COMPLEMENTARY ANGLE FORMULAS: The COMPLEMENTARY ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. 𝜋 𝜋 cos 𝜃 = sin (𝜃 + ) = − sin (𝜃 − ) 2 2 𝜋 𝜋 sin 𝜃 = cos (𝜃 − ) = − cos (𝜃 + ) 2 2 DOUBLE-ANGLE FORMULAS: The DOUBLE-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. sin 2𝛼 = 2 sin 𝛼 cos 𝛼 cos 2𝛼 = cos 2 𝛼 − sin2 𝛼 = 1 − 2sin2 𝛼 = 2 cos 2 𝛼 − 1
tan 2𝛼 =
(2 tan 𝛼) (1 − tan2 𝛼)
(cot 2 𝛼 − 1) cot 2𝛼 = (2 cot 𝛼)
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The solution to the trigonometric equation: 2 cos 𝑥 + sin 2𝑥 = 0 is most close to: 𝜋 7𝜋
A. 𝑥 = , 2
B. 𝑥 = 𝜋,
2 3𝜋 2
𝜋 3𝜋
C. 𝑥 = 2 ,
2
𝜋 3𝜋
D. 𝑥 = 4 ,
2
SOLUTION: In this problem, we are given the equation: 2 cos 𝑥 + sin 2𝑥 = 0 Right off the bat, we notice that our second term is a double angle, let’s simplify this. The DOUBLE-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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Specifically we are interested in the SINE TERM representation of a DOUBLE-ANGLE FORMULA, which states: sin 2𝛼 = 2 sin 𝛼 cos 𝛼 Using this formula, we can substitute and re-write the equation to read: 2 cos 𝑥 + 2 sin 𝑥 cos 𝑥 = 0 With cos x present in both terms, let’s factor it out, such that: 2 cos 𝑥 (1 + sin 𝑥) = 0 Let’s now solve for the roots starting with our COSINE TERM, given: 2 cos 𝑥 = 0 This expression is TRUE when:
𝑥=
𝜋 3𝜋 , 2 2
Now for our SINE term, given: 1 + sin 𝑥 = 0
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This expression is TRUE when:
𝑥=
3𝜋 2
NOTE: It is important to remember that trig functions typically have solutions in a range between 0 and 2𝜋. 𝝅 𝟑𝝅
The correct answer choice is C. 𝒙 = , 𝟐
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HALF-ANGLE FORMULAS: The HALF-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. They are stated as:
𝛼 1 − cos 𝛼 sin ( ) = ± √ 2 2
𝛼 1 + cos 𝛼 cos ( ) = ± √ 2 2
𝛼 1 − cos 𝛼 tan ( ) = ± √ 2 1 + cos 𝛼
𝛼 1 + cos 𝛼 cot ( ) = ± √ 2 1 − cos 𝛼
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
𝑥
The solutions of 2 − sin2 𝑥 = 2 cos 2 2 in the interval (0,2𝜋) are most close to: A. 𝑥 = 0, 𝑥 = 𝜋
B. 𝑥 = 4 , 𝑥 = 𝜋
𝜋 2 𝜋 2
C. 𝑥 = 2 , 𝑥 =
3𝜋
D. 𝑥 = 𝜋, 𝑥 =
𝜋
2 2
SOLUTION: We are given the equation:
2 − sin2 𝑥 = 2 cos 2
𝑥 2
We notice right off the bat that we are dealing with a half angle on the right side of this equation, let’s simply that. The HALF-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with
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In this particular problem, we are most concerned with the COSINE HALF-ANGLE FORMULA, which states:
𝛼 1 + cos 𝛼 cos ( ) = ± √ 2 2
Using this half-angle formula, we can re-write the cosine term such that: 2
2 − sin2 𝑥 = 2 (±√
1 + cos 𝑥 ) 2
Having both a square root and an squared term on the right hand side of the equation, they simply cancel themselves out, simplifying our expression as:
2 − sin2 𝑥 = 2 (
1 + cos 𝑥 ) 2
Staying on the right side, we can also cancel out the variable 2 in front of the term and in the denominator: 2 − sin2 𝑥 = 1 + cos 𝑥 Now let’s get rid of that squared SINE term using the PYTHAGOREAN IDENTITIES, which tell us: sin2 𝜃 + cos 2 𝜃 = 1 Made with
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And: sin2 𝜃 = 1 − cos 2 𝜃 We substitute this term in to our expression to get: 2 − (1 − cos 2 𝑥) = 1 + cos 𝑥 We can then simplify the expression: cos 2 𝑥 − cos 𝑥 = 0 Let’s now factor the expression as: cos 𝑥 (cos 𝑥 − 1) = 0 Solve for the value of “𝑥” for each factored expression to determine the solutions, which gives us: cos 𝑥 = 0
𝑥=
𝜋 3𝜋 , 2 2 𝝅
The correct answer choice is B. 𝒙 = 𝟐 , 𝒙 =
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SUM AND DIFFERENCE FORMULAS: The SUM AND DIFFERENCE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. sin(𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 sin(𝛼 − 𝛽) = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽 cos(𝛼 + 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽
tan(𝛼 + 𝛽) =
(tan 𝛼 + tan 𝛽) (1 − tan 𝛼 tan 𝛽)
tan(𝛼 − 𝛽) =
(tan 𝛼 − tan 𝛽) (1 + tan 𝛼 tan 𝛽)
cot(𝛼 + 𝛽) =
(cot 𝛼 cot 𝛽 − 1) (cot 𝛼 + cot 𝛽)
cot(𝛼 − 𝛽) =
(cot 𝛼 cot 𝛽 + 1) (cot 𝛽 − cot 𝛼)
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OTHER TRIGONOMETRIC IDENTITIES THAT ARE PROVIDED IN THE NCEES REFERENCE HANDBOOK:
sin 𝛼 sin 𝛽 =
1 [cos(𝛼 − 𝛽) − cos(𝛼 + 𝛽)] 2
cos 𝛼 cos 𝛽 =
1 [cos(𝛼 − 𝛽) + cos(𝛼 + 𝛽)] 2
sin 𝛼 cos 𝛽 =
1 [sin(𝛼 + 𝛽) + sin(𝛼 − 𝛽)] 2
1 1 sin 𝛼 + sin 𝛽 = 2 sin [ (𝛼 + 𝛽)] cos [ (𝛼 − 𝛽)] 2 2 1 1 sin 𝛼 − sin 𝛽 = 2 cos [ (𝛼 + 𝛽)] sin [ (𝛼 − 𝛽)] 2 2 1 1 cos 𝛼 + cos 𝛽 = 2 cos [ (𝛼 + 𝛽)] cos [ (𝛼 − 𝛽)] 2 2 1 1 cos 𝛼 − cos 𝛽 = −2 sin [ (𝛼 + 𝛽)] sin [ (𝛼 − 𝛽)] 2 2
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Simplify the expression: cos(𝑥 + 𝑦) + cos(𝑥 − 𝑦) sin 𝑥 cos 𝑦 A. 3 sin 𝑥 B. tan 3𝑥 C. 2 cot 𝑥 D. 5 sec 𝑥
SOLUTION: Right up front, we notice that we are working with a few trigonometric terms that fall use the SUM and DIFFERENCE of a pair of values. The SUM AND DIFFERENCE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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We are given the expression: cos(𝑥 + 𝑦) + cos(𝑥 − 𝑦) sin 𝑥 cos 𝑦 We will use the sum and difference formulas to expand out the COSINE terms in the numerator, specifically we will use: cos(𝛼 + 𝛽 ) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽 Plugging these IDENTITIES in to the original expression, we get: (cos 𝑥 cos 𝑦 + sin 𝑥 sin 𝑦) + (cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦) sin 𝑥 cos 𝑦 This creates a really nice expression in the Numerator that we can simplify. Combining like terms, we get: 2 cos 𝑥 cos 𝑦 sin 𝑥 cos 𝑦 Cancelling out the terms in both the Numerator and Denominator, we have: 2 cos 𝑥 sin 𝑥 Made with
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The GENERAL FORMULAS for the RATIO IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Using the ratio identity, we come to the final answer: 2 cos 𝑥 = 2 cot 𝑥 sin 𝑥 The correct answer choice is C. 𝟐 𝒄𝒐𝒕 𝒙
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CONCEPT INTRO: An IDENTITY is an expression or formula that is always true. Identities are used to solve equations as they can be multiplied on both sides of the equation. A TRIGONOMETRIC IDENTITY is an equation involving trigonometric functions that can be solved by any angle. Trigonometric identities have less to do with evaluating functions at specific angles and more to do with relationships between functions. During the exam, all of the TRIGONOMETRIC IDENTITIES you will ever need can be found in the NCEES SUPPLIED REFERENCE HANDBOOK. There are some general steps that can be followed when solving equations that use trigonometric identities: STEP 1: • Look for obvious simplifications. This should be your first step in solving any equation. Made with
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• If you can easily see one of the identities right away, go ahead and start substituting the identity into the equation to start simplifying terms. • If you have an expression with many different trig functions inside, try to think if it can be simplified in any other method obeying the laws of algebraic expressions. For example, can you use the conjugate or another identity to simplify terms. STEP 2: • Convert as many terms as you can to sine and cosine. You can do this using reciprocal identities, tangent identities, and cotangent identities. • Sine and cosine terms are easier to work with, and have identities that are easier to identify when being applied. • See if you can now combine like terms, eliminate fractions, and simplify the expressions on each side of the equation. • Try to identify what terms or expressions can be factored out and replaced with trigonometric identities.
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STEP 3: • Go through all of the trigonometric identities, and see if you can apply any of them. Pay attention to how terms are multiplied, what power terms are raised to, and the +/- of each term to figure out the best trigonometric identity to use. • The most commonly use trigonometric identities on the exam are the PYTHOGREAN IDENTITIES. STEP 4: • The final step is to make sure your answer is one of the answer choices. It is common to do everything correctly, and get your answer in a different form than that of one of the answer choices. • Don’t worry! You have the right answer, you may just need to apply another identity, or re-arrange your answer. • If you get to this point, and still can’t get an answer choice, convert everything to sine and cosine terms again. RECIPROCAL IDENTITIES: The GENERAL FORMULAS highlighting the RECIPROCAL IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with
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The can be stated as:
sin 𝜃 =
1 csc 𝜃
cos 𝜃 =
1 sec 𝜃
tan 𝜃 =
1 cot 𝜃
csc 𝜃 =
1 sin 𝜃
sec 𝜃 =
1 cos 𝜃
cot 𝜃 =
1 tan 𝜃
RATIO IDENTITIES: The FORMULAS for the RATIO IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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They can be stated as:
tan 𝜃 =
sin 𝜃 cos 𝜃
cot 𝜃 =
cos 𝜃 sin 𝜃
PYTHAGOREAN IDENTITIES: The FORMULAS for the PYTHAGOREAN IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. They are stated as: sin2 𝜃 + cos 2 𝜃 = 1 tan2 𝜃 + 1 = sec 2 𝜃 cot 2 𝜃 + 1 = csc 2 𝜃
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Perform the addition and simplification of the following trigonometric expression: sin 𝜃 cos 𝜃 + 1 + cos 𝜃 sin 𝜃 A. tan 𝜃 B. csc 𝜃 C. cot 𝜃 D. sin 𝜃
SOLUTION: We are given the expression originally as: sin 𝜃 cos 𝜃 + 1 + cos 𝜃 sin 𝜃 We now need to look for ways to expand, rearrange, combine…whatever we can do to simplify it the best that we can.
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Establishing the common denominator, we are able to pull these two terms together such that: sin 𝜃 cos 𝜃 (sin 𝜃)(sin 𝜃) + (cos 𝜃)(1 + cos 𝜃) + = 1 + cos 𝜃 sin 𝜃 (1 + cos 𝜃)(sin 𝜃) The sine and cosine terms are now beginning to reflect a trigonometric identity, which is what we want in route to simplifying. Multiplying out the factored terms in the numerator, we get: sin2 𝜃 + cos 2 𝜃 + cos 𝜃 (1 + cos 𝜃)(sin 𝜃) The most commonly used trigonometric identities that we will encounter on the exam are those of the PYTHOGREAN IDENTITIES. The FORMULAS for the PYTHAGOREAN IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The PYTHAGOREAN IDENTITIES state that: sin2 𝜃 + cos 2 𝜃 = 1 tan2 𝜃 + 1 = sec 2 𝜃
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cot 2 𝜃 + 1 = csc 2 𝜃 Heading back to the expression we have at this point, we can simplify some terms into a Pythagorean identity, such that: sin2 𝜃 + cos 2 𝜃 + cos 𝜃 (1 + cos 𝜃) = (1 + cos 𝜃)(sin 𝜃) (1 + cos 𝜃)(sin 𝜃) With both (1 + cos 𝜃) in the numerator and the denominator, let’s divide it out, leaving us with: (1 + cos 𝜃) 1 = (1 + cos 𝜃)(sin 𝜃) sin 𝜃 Using the reciprocal identity, our final answer is: 1 = csc 𝜃 sin 𝜃 The correct answer choice is B. 𝒄𝒔𝒄 𝜽
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COMPLEMENTARY ANGLE FORMULAS: The COMPLEMENTARY ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. 𝜋 𝜋 cos 𝜃 = sin (𝜃 + ) = − sin (𝜃 − ) 2 2 𝜋 𝜋 sin 𝜃 = cos (𝜃 − ) = − cos (𝜃 + ) 2 2 DOUBLE-ANGLE FORMULAS: The DOUBLE-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. sin 2𝛼 = 2 sin 𝛼 cos 𝛼 cos 2𝛼 = cos 2 𝛼 − sin2 𝛼 = 1 − 2sin2 𝛼 = 2 cos 2 𝛼 − 1
tan 2𝛼 =
(2 tan 𝛼) (1 − tan2 𝛼)
(cot 2 𝛼 − 1) cot 2𝛼 = (2 cot 𝛼)
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The solution to the trigonometric equation: 2 cos 𝑥 + sin 2𝑥 = 0 is most close to: 𝜋 7𝜋
A. 𝑥 = , 2
B. 𝑥 = 𝜋,
2 3𝜋 2
𝜋 3𝜋
C. 𝑥 = 2 ,
2
𝜋 3𝜋
D. 𝑥 = 4 ,
2
SOLUTION: In this problem, we are given the equation: 2 cos 𝑥 + sin 2𝑥 = 0 Right off the bat, we notice that our second term is a double angle, let’s simplify this. The DOUBLE-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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Specifically we are interested in the SINE TERM representation of a DOUBLE-ANGLE FORMULA, which states: sin 2𝛼 = 2 sin 𝛼 cos 𝛼 Using this formula, we can substitute and re-write the equation to read: 2 cos 𝑥 + 2 sin 𝑥 cos 𝑥 = 0 With cos x present in both terms, let’s factor it out, such that: 2 cos 𝑥 (1 + sin 𝑥) = 0 Let’s now solve for the roots starting with our COSINE TERM, given: 2 cos 𝑥 = 0 This expression is TRUE when:
𝑥=
𝜋 3𝜋 , 2 2
Now for our SINE term, given: 1 + sin 𝑥 = 0
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This expression is TRUE when:
𝑥=
3𝜋 2
NOTE: It is important to remember that trig functions typically have solutions in a range between 0 and 2𝜋. 𝝅 𝟑𝝅
The correct answer choice is C. 𝒙 = , 𝟐
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HALF-ANGLE FORMULAS: The HALF-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. They are stated as:
𝛼 1 − cos 𝛼 sin ( ) = ± √ 2 2
𝛼 1 + cos 𝛼 cos ( ) = ± √ 2 2
𝛼 1 − cos 𝛼 tan ( ) = ± √ 2 1 + cos 𝛼
𝛼 1 + cos 𝛼 cot ( ) = ± √ 2 1 − cos 𝛼
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
𝑥
The solutions of 2 − sin2 𝑥 = 2 cos 2 2 in the interval (0,2𝜋) are most close to: A. 𝑥 = 0, 𝑥 = 𝜋
B. 𝑥 = 4 , 𝑥 = 𝜋
𝜋 2 𝜋 2
C. 𝑥 = 2 , 𝑥 =
3𝜋
D. 𝑥 = 𝜋, 𝑥 =
𝜋
2 2
SOLUTION: We are given the equation:
2 − sin2 𝑥 = 2 cos 2
𝑥 2
We notice right off the bat that we are dealing with a half angle on the right side of this equation, let’s simply that. The HALF-ANGLE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with
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In this particular problem, we are most concerned with the COSINE HALF-ANGLE FORMULA, which states:
𝛼 1 + cos 𝛼 cos ( ) = ± √ 2 2
Using this half-angle formula, we can re-write the cosine term such that: 2
2 − sin2 𝑥 = 2 (±√
1 + cos 𝑥 ) 2
Having both a square root and an squared term on the right hand side of the equation, they simply cancel themselves out, simplifying our expression as:
2 − sin2 𝑥 = 2 (
1 + cos 𝑥 ) 2
Staying on the right side, we can also cancel out the variable 2 in front of the term and in the denominator: 2 − sin2 𝑥 = 1 + cos 𝑥 Now let’s get rid of that squared SINE term using the PYTHAGOREAN IDENTITIES, which tell us: sin2 𝜃 + cos 2 𝜃 = 1 Made with
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And: sin2 𝜃 = 1 − cos 2 𝜃 We substitute this term in to our expression to get: 2 − (1 − cos 2 𝑥) = 1 + cos 𝑥 We can then simplify the expression: cos 2 𝑥 − cos 𝑥 = 0 Let’s now factor the expression as: cos 𝑥 (cos 𝑥 − 1) = 0 Solve for the value of “𝑥” for each factored expression to determine the solutions, which gives us: cos 𝑥 = 0
𝑥=
𝜋 3𝜋 , 2 2 𝝅
The correct answer choice is B. 𝒙 = 𝟐 , 𝒙 =
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𝟑𝝅 𝟐
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SUM AND DIFFERENCE FORMULAS: The SUM AND DIFFERENCE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. sin(𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 sin(𝛼 − 𝛽) = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽 cos(𝛼 + 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽
tan(𝛼 + 𝛽) =
(tan 𝛼 + tan 𝛽) (1 − tan 𝛼 tan 𝛽)
tan(𝛼 − 𝛽) =
(tan 𝛼 − tan 𝛽) (1 + tan 𝛼 tan 𝛽)
cot(𝛼 + 𝛽) =
(cot 𝛼 cot 𝛽 − 1) (cot 𝛼 + cot 𝛽)
cot(𝛼 − 𝛽) =
(cot 𝛼 cot 𝛽 + 1) (cot 𝛽 − cot 𝛼)
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OTHER TRIGONOMETRIC IDENTITIES THAT ARE PROVIDED IN THE NCEES REFERENCE HANDBOOK:
sin 𝛼 sin 𝛽 =
1 [cos(𝛼 − 𝛽) − cos(𝛼 + 𝛽)] 2
cos 𝛼 cos 𝛽 =
1 [cos(𝛼 − 𝛽) + cos(𝛼 + 𝛽)] 2
sin 𝛼 cos 𝛽 =
1 [sin(𝛼 + 𝛽) + sin(𝛼 − 𝛽)] 2
1 1 sin 𝛼 + sin 𝛽 = 2 sin [ (𝛼 + 𝛽)] cos [ (𝛼 − 𝛽)] 2 2 1 1 sin 𝛼 − sin 𝛽 = 2 cos [ (𝛼 + 𝛽)] sin [ (𝛼 − 𝛽)] 2 2 1 1 cos 𝛼 + cos 𝛽 = 2 cos [ (𝛼 + 𝛽)] cos [ (𝛼 − 𝛽)] 2 2 1 1 cos 𝛼 − cos 𝛽 = −2 sin [ (𝛼 + 𝛽)] sin [ (𝛼 − 𝛽)] 2 2
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TRIGONOMETRIC IDENTITIES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Simplify the expression: cos(𝑥 + 𝑦) + cos(𝑥 − 𝑦) sin 𝑥 cos 𝑦 A. 3 sin 𝑥 B. tan 3𝑥 C. 2 cot 𝑥 D. 5 sec 𝑥
SOLUTION: Right up front, we notice that we are working with a few trigonometric terms that fall use the SUM and DIFFERENCE of a pair of values. The SUM AND DIFFERENCE FORMULAS can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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We are given the expression: cos(𝑥 + 𝑦) + cos(𝑥 − 𝑦) sin 𝑥 cos 𝑦 We will use the sum and difference formulas to expand out the COSINE terms in the numerator, specifically we will use: cos(𝛼 + 𝛽 ) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽 Plugging these IDENTITIES in to the original expression, we get: (cos 𝑥 cos 𝑦 + sin 𝑥 sin 𝑦) + (cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦) sin 𝑥 cos 𝑦 This creates a really nice expression in the Numerator that we can simplify. Combining like terms, we get: 2 cos 𝑥 cos 𝑦 sin 𝑥 cos 𝑦 Cancelling out the terms in both the Numerator and Denominator, we have: 2 cos 𝑥 sin 𝑥 Made with
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The GENERAL FORMULAS for the RATIO IDENTITIES can be referenced under the topic of IDENTITIES on page 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Using the ratio identity, we come to the final answer: 2 cos 𝑥 = 2 cot 𝑥 sin 𝑥 The correct answer choice is C. 𝟐 𝒄𝒐𝒕 𝒙
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