Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

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1 Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the terminal side of θ and the unit circle we have that sinθ) is defined to be y, while cscθ) is defined to be y. Therefore sinθ) = cscθ).. State the definition of an odd Function and use it to determine if fθ) = sinθ)tanθ) is odd, even or neither. a) Definition: Solution: fθ) is an odd function if and only if f θ) = fθ) for all θ in the domain. b) Odd, Even or Neither Justification of fθ): Solution: Since f θ) = sin θ) tan θ) = sinθ)) tanθ)) = sinθ) tanθ) = fθ), we have that fθ) is an even function. 3. State a FTI identity and then verify it. a) Identity: Solution: tan θ) + = sec θ) is one identity. b) Verification: Solution: The FTI says that sin θ) + cos θ) =, therefore sin θ) cos θ) + = sec θ) and hence tan θ) + = sec θ) as desired. 4. State a Graph identity and then verify it with a step by step graph approach. sin cos θ) + cos θ) ) = θ) cos θ) or that a) Identity: Solution: One identity is cos π θ) = sinθ). b) Verification: Solution: Observe that cos π θ) = cos θ π )) = cosθ π ) since cosine is even. Now since the graph of fθ) = cosθ) is and the graph of fθ π is

2 which is the same as the graph of sine. Therefore cos π θ) = sinθ). 5. State a Double Angle ID and then verify it. a) Identity: Solution: One is cosθ) = cos θ) sin θ). b) Verification: Solution: Observe that cosθ) = cosθ + θ) and by the angle sum id we have cosθ) = cosθ) cosθ) sinθ) sinθ) = cos θ) sin θ) as desired. 6. Verify the ID sinθ) cosβ) = sinθ + β) + sinθ β)). a) Verification: Solution: Observe that and thus sinθ + β) + sinθ β)) = [sinθ) cosβ) + sinβ) cosθ)) + sinθ) cosβ) sinβ) cosθ))] sinθ + β) + sinθ β)) = [ sinθ) cosβ)] = sinθ) cosβ)

3 Evaluate, Simplify and Solve Questions 7. If cosθ) = x for some positive real number x < with π θ π evaluate cot θ) in terms of x. Draw a picture to justify your work. Solution: Since cosine is positive in quadrants and 4 but we are restricted to quadrants, and 3 we know that θ must terminate in quadrant. Since cosine is adjacent over hypotenuse we can now plot the angle and draw the reference triangle as follows using the Pythagorean theorem to find the missing side: x Now since cotangent is odd we have that cot θ) = cotθ) =. x 8. Simplify the expression 4 tan θ) sin θ) sin θ) cos θ π ) π ) + cos θ) sec θ Solution: 4 tan θ) sin θ) cos θ π ) + π ) 4 tanθ) sinθ) + cos θ) = 6 6 sin θ) sec θ + sinθ) 6 sin θ) cscθ) + cosθ) By using the Odd and Even Function Prop and Graph IDS = By using the FTI, Basic IDs and double angle ID 4 sinθ) sinθ) 4 cosθ) cos θ) + sinθ) sinθ) + cos θ) sin θ) By simplifying and using Basic IDs = = tan θ) + sin θ) + cos θ) tan θ) + with FTI = sec θ) = secθ) 3

4 9. Solve for all values of θ such that a) cosθ) = cos θ) +. Solution: Using the double angle ID for cosine we have that cosθ) = cos θ) + cos θ) sin θ) = cos θ) + This results in sin θ) = which has no solution. Thus θ =. b) Hint: Rewrite 3θ as θ + θ and use an angle sum identity. Solution: Observe that by sum angle ID sin3θ) cscθ) = sin3θ) cscθ) = sinθ + θ) cscθ) = [sinθ) cosθ) + sinθ) cosθ] sinθ) = By double angle ID by Simplifying [ sinθ) [ cos θ) sin θ) ] + [ sinθ) cosθ)] cosθ) ] sinθ) = cos θ) sin θ) + cos θ) = 4 cos θ) = cosθ) = ± 4 = ± θ = ± π 6 + πk, k Z 4

5 0. Evaluate the following using ANGLE SUM IDENTITIES: ) 5π a) cos ) 5π 4π Solution: Since cos = cos 3 + 3π ) we have by the angle sum ID that 4 cos b) tan 3π ) Solution: Since tan ) 5π = cos 4π ) ) 3 ) cos3π 4 ) sin4π 3 ) sin3π 4 ) = ) ) 3 3π ) 7π = tan π ) we have by the angle sum ID that 3 4 tan 3π ) tan 7π = 3 ) tanπ 4 ) 3 + tan 7π = 3 ) tanπ 4 ) 3. Evaluate the following using HALF ANGLE IDENTITIES: ) 3π a) sin Solution: Since 3π is in the third quadrant we have that sin we have that ) 3π sin = b) cot 7π ) 8 cos 3π 6 ) = ) 3π is negative. Thus by the half angle ID 3 Solution: Since 7π 8 is in quadrant 3 we have that cotangent is positive. Thus by half angle ID we have cot 7π ) + cos 7π 4 = ) 8 cos 7π 4 ) = + 5

6 . Use the bank of functions below to identify the graphs that follow, a) f θ) = sinθ) + cos3θ), g θ) = sinθ) + cos5θ), h θ) = sinθ) + cos7θ) f θ) = sin3θ) + cosθ), g θ) = sin5θ) + cosθ), h θ) = sin7θ) + cosθ) f 3 θ) = θ + tanθ), g 3 θ) = θ + tanθ), h 3θ) = θ + tan θ) θ + tan θ) 6

7 b) sinθ) + cos5θ) c) sinθ) + cos7θ) 7

8 3. For the function fθ) = cosθ) + cos4θ): a) Find the period of fθ) Solution: The period of cosθ) is π fθ) = LCMπ, π ) = π. = π, while the period of cos4θ) is π 4 = π. Therefore the period of b) Solve for all θ such that fθ) = 0. Hint: Use a sum to product ID) Solution: Using the sum to product ID we have that Thus fθ) = 0 implies cos3θ) cosθ) = 0 or that fθ) = cosθ) + cos4θ) = cos 6θ ) cos θ ) = cos3θ) cos θ) = cos3θ) cosθ) cos3θ) = 0 or cosθ) = 0 This results in 3θ = π + πk or θ = π + πk for k Z. This results in θ = π 6 + π 3 k and π + πk. 4. Simplify. Hint: Be sure to use an angle sum ID and a product to sum ID. Solution: Observe that sinθ) + sin3θ)) cscθ) secθ cos3θ) cosθ) + cos4θ) cscθ) secθ sinθ) + sin3θ)) cos3θ) cosθ) + cos4θ) [ ] = [ sinθ) cos θ)] sinθ) cosθ) cos4θ) + cosθ) Using the sum to product ID and the prod to sum ID + cos4θ) = sinθ) cos θ) sinθ) cosθ) cos4θ) cosθ) + cos4θ) by using the double angle ID, the Even function ID and simplyfying = cosθ) cosθ) 8