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Trigonometry II Trigonometry II© Standards ANSWERS Copyright 2006,Test Barry Mabillard. 0 www.math40s.com www.math40s.com 2 4 1. If cos α = − , where sin α > 0 , and sin β = , where tan β > 0 , determine the exact 5 9 value of cos (α + β ) First determine the quadrant α is in: cos α < 0 sin α > 0 Then determine the quadrant β is in: sin β > 0 tan β > 0 Now solve the triangles for both α and β; use Pythagoras to find the unknown sides. a 2 + b2 = c2 a 2 + b2 = c 2 ( −2 ) ( 4) 2 + b 2 = 52 2 + b 2 = 92 4 + b 2 = 25 16 + b 2 = 81 b 2 = 21 b 2 = 65 b = 21 b = 65 At this point, state all required trigonometric ratios: 2 65 cos β = cos α = − 9 5 4 21 sin β = sin α = 9 5 Finally, evaluate cos (α + β ) cos (α + β ) = cos α cos β − sin α sin β ⎛ 2 ⎞ ⎛ 65 ⎞ ⎛ 21 ⎞ ⎛ 4 ⎞ cos (α + β ) = ⎜ − ⎟ ⎜⎜ ⎟−⎜ ⎟⎜ ⎟ ⎝ 5 ⎠ ⎝ 9 ⎠⎟ ⎝⎜ 5 ⎠⎟ ⎝ 9 ⎠ 2 65 4 21 − 45 45 2 65 - 4 21 cos (α + β ) = 45 cos (α + β ) = Trigonometry II Standards Test ANSWERS 1 www.math40s.com 2. Express 1 − sin 2 (α + β ) using only cosine. Start with the identity: sin 2 θ + cos 2 θ = 1 Rearrange to get: cos 2 θ = 1 − sin 2 θ It follows that: cos 2 (α + β ) = 1 − sin 2 (α + β ) The answer is cos 2 (α + β ) 3. If cos 3θ = 3 , where 3θ is an acute angle, determine the exact value of θ. 2 3 π 11π , you get x = , If you evaluate cosx = . 6 6 2 π is acute (<90°). 6 π π 1 π Divide by 3 to get the angle θ: ÷ 3 = × = 6 6 3 18 You do not want to use any other angles since only 3 in your calculator using 2 degree mode, then convert the answer to a radian fraction. *Alternatively, you could graph each side of cos 3θ = 4. Determine the exact value of tan (15o ) Rewrite as tan (45 o - 30 o ) and use the formula tan (α − β ) = tan 45o − tan 300 1 + tan 45o tan 300 1 1 1− 1− 3 3 = = 1 ⎛ 1 ⎞ 1+ 1 + (1) ⎜ ⎟ 3 ⎝ 3⎠ tan α − tan β 1 + tan α tan β tan ( 45o − 300 ) = = 3 − 3 3 + 3 1 3 1 3 = 3 −1 3 3 +1 3 = 3 −1 3 × 3 3 +1 = 3 -1 3 +1 Get a common denominator for top & bottom Evaluate the required tan ratios before doing the calculation on the left. 1 o sin 30 1 2 1 tan 30o = = 2 = × = o cos 30 3 2 3 3 2 3 o sin 45 tan 45o = = 2 =1 o cos 45 3 2 Divide fractions by multiplying the reciprocal Trigonometry II Standards Test ANSWERS 2 www.math40s.com 5. Prove the identity: 1 − cot 2 x = sin 2 x − cos 2 x 1 + cot 2 x 1- cot 2 x 1+ cot 2 x 1- cot 2 x = csc 2 x = (1- cot 2 x ) sin 2 x ⎛ cos 2 x ⎞ 2 = ⎜1⎟ sin x 2 sin x ⎝ ⎠ 2 = sin x - cos 2 x 6. Solve for x, where 0 ≤ x ≤ 2π : csc2 x − csc x = 2 Rewrite as csc 2 x - cscx - 2 = 0 Then factor: (cscx - 2 )(cscx + 1 ) = 0 Now equate each set of brackets to zero and solve for x. cscx - 2 = 0 cscx = 2 1 2 π 5π x= , 6 6 sinx = cscx + 1= 0 cscx = -1 sinx = -1 3π x= 2 The answer is x = π 5 π 3π 6 , 6 , 2 7. Solve for x: 2 cos 2 x + cos x − 1 = 0 Factor to obtain: (2cosx - 1 )(cosx + 1 ) = 0 Now equate each set of brackets to zero and solve for x. 2cosx - 1= 0 2cosx = 1 1 2 π 5π x= , 3 3 cosx = cosx + 1= 0 cosx = -1 x=π The answer is x = Trigonometry II Standards Test ANSWERS π 3 ,π , 3 5π 3 www.math40s.com 8. Simplify: sin x cot x + cos x sin x ⎛ cosx ⎞ sinx ⎜ ⎟+ cosx sinx ⎠ ⎝ = sinx cosx + cosx = sinx 2cosx = sinx = 2cotx 9. Solve ( 2 cos θ + 1) ( tan θ − 1) = 0 for θ in the interval π ≤ x ≤ 2π 2 Equate each set of brackets to zero and solve for θ. 2cosθ + 1= 0 2cosθ = -1 tanθ - 1= 0 tanθ = 1 π 5π θ= , 4 4 1 2 2 π 4π θ= , 3 3 cosθ = - θ= 2 π 5 π 4π , , 3 4 3 (Omit π 4 since it lies outside the specified domain.) 10. Determine the exact value of 11π a) cos 12 11π First convert to degrees. This will make 12 b) sec sec165 o = the remainder of the calculation easier than working in radian fractions. ( ) 11π 180 o × = 165 o π 12 Write as cos 120 o + 45 o , then apply the 1 cos165 o This is simply the reciprocal of the answer you found in part a) The answer is cosine sum formula ( 11π 12 ) 4 - 2- 6 o o o o o o cos 120 + 45 = cos120 cos45 - sin120 cos45 ⎛ 1 ⎞⎛ 2 ⎞ ⎛ 3 ⎞⎛ 2 ⎞ = ⎜- ⎟ ⎜⎜ ⎟- ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ - 2 6 4 4 - 2- 6 = 4 = Trigonometry II Standards Test ANSWERS 4 www.math40s.com 11. Prove the identity 12. If tan θ = − cot θ sec 2 θ = tan θ cot 2 θ + 1 cotθsec 2θ cot 2θ + 1 cotθsec 2θ = csc 2θ 1 2 = cotθ • cos θ 1 sin 2θ cosθ ⎛ 1 ⎞ 2 = ⎜ ⎟ ( sin θ ) sinθ ⎝ cos 2θ ⎠ sinθ = cosθ = tanθ 3 3π and < θ < 2π , state the exact value of cos 2θ 2 2 Draw the triangle and find the hypotenuse using Pythagoras: a 2 + b2 = c2 ( 2 ) + ( −3) 2 2 = c2 4 + 9 = c2 13 = c 2 c = 13 From the diagram, cosθ = 2 13 Now use the formula: cos2θ = 2cos 2θ - 1 2 ⎛ 2 ⎞ cos2θ = 2 ⎜ ⎟ -1 ⎝ 13 ⎠ ⎛ 4 ⎞ cos2θ = 2 ⎜ ⎟- 1 ⎝ 13 ⎠ 8 cos2θ = -1 13 5 cos2θ = 13 Trigonometry II Standards Test ANSWERS 5 www.math40s.com 3 2 To solve this equation algebraically, you need to determine the solutions within 2 rotations of the unit circle. 3 π 2π Solve the equation sinx = to get the angles , and 2 3 3 both of which are within the first rotation. 13. Solve for θ over [ 0, 2π ] : sin 2θ = Add the period to each one to get the co-terminal angles in the second rotation. π π 6π 7 π + 2π = + = 3 3 3 3 2π 2π 6π 8π + 2π = + = 3 3 3 3 Finally, take all your solutions π 2π 7 π 8π and divide by 2. , , 3 3 3 π 2π 7 π 8π The answer is , , , 6 6 6 6 3 , 14. Solve the following equation for x ∈ R : 2 cos 2θ + 1 = 0 Get cos2θ by itself before trying to determine the angles: 2cos2θ = -1 -1 cos2θ = 2 To solve this equation algebraically, you need to determine the solutions within 2 rotations of the unit circle. 1 2π 4π to get the angles , Solve the equation cosx = and 2 3 3 both of which are within the first rotation. Add the period to each one to get the co-terminal angles in the second rotation. 2π 2π 6π 8π + 2π = + = 3 3 3 3 4π 4 π 6 π 10 π π + 2π = + = 3 3 3 3 2 π 4 π 8 π 10 π and divide by 2. , , , 3 3 3 3 2 π 4 π 8 π 10 π , , , The answer is 6 6 6 6 Finally, take all your solutions Trigonometry II Standards Test ANSWERS 6 www.math40s.com 15. If α and β are second quadrant angles, and cos α = − determine the exact value of sin (α − β ) 2 1 and sin β = , 3 4 Solve the triangles for both α and β; use Pythagoras to find the unknown sides. a 2 + b2 = c2 a 2 + b2 = c 2 ( −2 ) 12 + b 2 = 42 + b 2 = 32 2 4 + b2 = 9 1 + b 2 = 16 b =5 b 2 = 15 b= 5 b = 15 2 At this point, state all required trigonometric ratios: 2 3 5 sin α = 3 cos α = − Use a negative since the b-value lies on the negative x-axis. − 15 4 1 sin β = 4 cos β = Finally, evaluate sin (α - β ) sin (α − β ) = sin α cos β − cos α sin β ⎛ 5 ⎞ ⎛ − 15 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ sin (α − β ) = ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ − ⎜ − ⎟ ⎜ ⎟ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ − 75 2 + 12 12 2 - 75 sin (α − β ) = 12 sin (α − β ) = 16. Solve for x over the interval [ 0, 2π ] for sin 2 x = sin x Set the equation to zero: sin 2 x - sinx = 0 Then factor: sinx ( sinx - 1 ) = 0 Now set each factor equal to zero and solve for the angles sinx = 0 x = 0, π ,2 π sinx - 1= 0 sinx = 1 x= π 2 The solution is π x=0, , π ,2π 2 Trigonometry II Standards Test ANSWERS 7 www.math40s.com 17. Prove 1 1 + = 2 csc θ 1 + cos θ 1 − cos θ 1 1 ⎛ ⎞ ⎛ 1- cosθ ⎞ ⎛ ⎞ ⎛ 1+ cosθ ⎞ ⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ⎟ ⎝ 1+ cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1+ cosθ ⎠ 1 1 ⎛ ⎞ ⎛ 1- cosθ ⎞ ⎛ ⎞ ⎛ 1+ cosθ ⎞ =⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ⎟ ⎝ 1+ cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1+ cosθ ⎠ 1- cosθ 1+ cosθ = + 2 1- cos θ 1- cos 2θ 1- cosθ + 1+ cosθ = 1- cos 2θ 2 = 1- cos 2θ 2 = sin 2θ = 2cscθ 18. Solve for x over the interval [ 0, 2π ] : 1 + tan 2 x = 3 . State the solutions as radians to three decimal places First isolate tanx : 1+ tan 2 x = 3 tan 2 x = 2 tanx = ± 2 Now solve in your calculator by graphing Y1 = tan x Y2 = + 2 Y3 = − 2 The answer is x = 0.9555, 2.186, 4.097, 5.328 Trigonometry II Standards Test ANSWERS 8 www.math40s.com 19. a) Prove sin 2 x = tan x 1 + cos 2 x sin2x 1+ cos2x 2sinxcosx = 1+ (2cos 2 x - 1 ) 2sinxcosx 2cos 2 x sinx = cosx = tanx = sin 2 x is undefined 1 + cos 2 x The expression is undefined when 1+ cos2x = 0 1+ cos2x = 0 b) State a value of x where cos2x = -1 First solve cosx = -1 over two rotations: The solution to this is x = π ,3 π Now divide by the 2 to get the solutions to cos2x = -1 π 3π The final answer is x = , 2 2 (You could also solve this by graphing Y1 = cos2x & Y2 = -1 in your calculator and finding the points of intersection) ⎛π ⎞ 20. Find the exact value of 2 cos 2 ⎜ ⎟ − 1 ⎝8⎠ Use the identity cos2α = 2cos 2 α - 1 ⎛π ⎞ cos2α = 2cos 2 ⎜ ⎟- 1 ⎝8 ⎠ What goes here is double what goes here. π 8 = π ⎛π = 2cos 2 ⎜ 4 ⎝8 π 2 The answer is: cos = 4 2 We now have the equation cos Trigonometry II Standards Test ANSWERS 9 π 2× 4 ⎞ ⎟- 1 ⎠ www.math40s.com ⎛ cos 2 x ⎞ + x sin ⎜ ⎟ sin x ⎠ 21. Simplify ⎝ csc x ⎛ cos 2 x ⎞ sinx + ⎜ ⎟ sinx ⎠ ⎝ cscx ⎛ cos 2 x = ⎜ sinx + sinx ⎝ ⎞ ⎟× sinx ⎠ = sin 2 x + cos 2 x =1 22. Find the exact value of: ⎛ 7π ⎞ a) cos ⎜ ⎟ ⎝ 12 ⎠ First express the angle in terms of degrees: 7 π 180 o × = 105 o π 12 cos (105 o ) = cos (60 o + 45 o ) cos (60 o + 45 o ) = cos60 o cos45 o - sin60 o sin45 o ⎛ 1 ⎞⎛ 2 ⎞ ⎛ 3 ⎞⎛ 2 ⎞ = ⎜ ⎟ ⎜⎜ ⎟- ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 6 4 4 2- 6 = 4 = ⎛ 7π ⎞ b) sec ⎜ ⎟ ⎝ 12 ⎠ This is simply the reciprocal of the answer you found in part a) 4 ⎛7 π ⎞ sec ⎜ ⎟= 2- 6 ⎝ 12 ⎠ Trigonometry II Standards Test ANSWERS 10 www.math40s.com 23. Express tan 2 θ using only sin θ sin 2θ cos 2θ Rewrite the identity sin 2θ + cos 2θ = 1 as cos 2θ = 1- sin 2θ Start with the identity tan 2θ = Finally, 24. Prove: sin 2θ sin 2θ = cos 2θ 1- sin 2θ cos x sin x + 1 + = 2sec x sin x + 1 cos x ⎛ cosx ⎞ ⎛ cosx ⎞ ⎛ sinx + 1 ⎞ ⎛ sinx + 1 ⎞ ⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ⎟ ⎝ sinx + 1 ⎠ ⎝ cosx ⎠ ⎝ cosx ⎠ ⎝ sinx + 1 ⎠ cos 2 x sin 2 x + 2sinx + 1 + cosx ( sinx + 1 ) cosx ( sinx + 1 ) cos 2 x + sin 2 x + 2sinx + 1 cosx ( sinx + 1 ) 1+ 2sinx + 1 cosx ( sinx + 1 ) 2 + 2sinx cosx ( sinx + 1 ) 2 (1+ sinx ) cosx ( sinx + 1 ) 2 cosx = 2secx = 25. Solve for θ in the interval [ 0, 2π ] : 2sin 2 θ − 3sin θ + 1 = 0 Factor to obtain: (2sinθ - 1 )( sinθ - 1 ) = 0 2sinθ - 1= 0 2sinθ = 1 1 sinθ = 2 π 5π θ= , 6 6 Now solve for the angles: sinθ - 1= 0 sinθ = 1 θ= π 2 The answer is θ = Trigonometry II Standards Test ANSWERS π π 5π , , 6 2 6 11 www.math40s.com 26. If cos θ = − 4 and csc θ > 0 , determine the exact value of: 5 a) tan 2θ First determine which quadrant the angle is found in: Now fill in the triangle and use Pythagoras to determine the unknown side: a 2 + b2 = c 2 ( −4 ) 2 + b 2 = 52 16 + b 2 = 25 b2 = 9 b=3 Use the formula tan2θ = ⎛ 3 2 ⎜⎝ 4 tan2θ = ⎛ 3 1- ⎜⎝ 4 6 4 = 9 116 6 4 = 16 9 16 16 6 = 4 7 16 6 16 =- × 4 7 24 =7 2tanθ 1- tan 2θ ⎞ ⎟ ⎠ 2 ⎞ ⎟ ⎠ From the triangle, 3 sinθ = 5 4 cosθ = 5 3 tanθ = 4 b) sin 2θ Use the formula sin2θ = 2sinθcosθ ⎛ 3 ⎞⎛ 4 ⎞ sin2θ = 2 ⎜ ⎟⎜- ⎟ ⎝ 5 ⎠⎝ 5 ⎠ 24 sin2θ = 25 Trigonometry II Standards Test ANSWERS 12 www.math40s.com 27. Solve the equation csc θ = −3 , where θ ∈ R . State your solution to three decimal places. Solve by graphing: The question specifies the domain as θ ∈ R , so a general solution is required. The (positive) solutions are 3.481 and 5.943 Since the solutions repeat themselves every period, write the general solution as: ⎧3.481+ 2k π , k ∈ I ⎫ θ= ⎨ ⎬ ⎩5.943+ 2k π , k ∈ I ⎭ 28. Show that sin 8x is equivalent to 2sin 4 x cos 4 x Start with the identity sin (2x ) = 2sinxcosx What goes here is double what goes here. In the equation sin8x = 2sin4xcos4x the 4x here becomes the 8x here 29. Express cot θ sec θ as a single trigonometric function cotθsecθ cosθ 1 = × sinθ cosθ 1 = sinθ = cscθ Trigonometry II Standards Test ANSWERS 13 www.math40s.com 30. If sin θ = 12 and cos θ < 0 , find the exact value of 13 a) tan θ First determine which quadrant the angle is found in: a2 + b2 = c2 (12 ) 2 + b 2 = 132 144 + b 2 = 169 b 2 = 65 b=5 From the triangle, the trig rations are: 12 sinθ = 13 5 cosθ = 13 12 tanθ = 5 The answer is tanθ = - π⎞ ⎛ b) cos ⎜ θ − ⎟ 4⎠ ⎝ 12 5 π⎞ ⎛ Expand cos ⎜θ - ⎟ using the subtraction law for cosine: ⎝ 4⎠ π⎞ π π ⎛ cos ⎜θ - ⎟ = cosθcos + sinθsin 4 4 ⎝ 4⎠ Now use the values found in part a) from the triangle, and values from the unit circle: π π ⎛ 5 ⎞ ⎛ 2 ⎞ ⎛ 12 ⎞ ⎛ 2 ⎞ cosθcos 4 + sinθsin 4 = ⎜⎟ ⎟+ ⎜ ⎟⎜ ⎟⎜ ⎝ 13 ⎠ ⎜⎝ 2 ⎟⎠ ⎝ 13 ⎠ ⎜⎝ 2 ⎟⎠ 5 2 12 2 + 26 26 12 2 - 5 2 = 26 =- Trigonometry II Standards Test ANSWERS 14 www.math40s.com 31. Find the exact values of x in the interval [ 0, 2π ] : cos 2 x + cos x + 1 = sin 2 x cos 2 x + cosx + 1= sin 2 x cos 2 x + cosx = sin 2 x - 1 cos 2 x + cosx = -cos 2 x 2cos 2 x + cosx = 0 2cosx + 1= 0 2cosx = -1 1 cosx = 2 2 π 4π x= , 3 3 cosx (2cosx + 1 ) = 0 Now solve for x: cosx = 0 π 3π x= , 2 2 The full solution set is x = 32. Prove the identity: π 2 π 4π 3π 2 , 3 , 3 , 2 cos x − cos3 x = cot x sin 3 x cosx - cos 3 x sin 3 x cosx (1- cos 2 x ) = sin 3 x cosxsin 2 x = sin 3 x cosx = sinx = cotx Trigonometry II Standards Test ANSWERS 15 www.math40s.com 33. Given the point shown on the circle, determine the value of tan 2β Recall that the x-coordinate is cosθ , and the y-coordinate is sinθ . 5 adj = 13 hyp 12 opp sinβ = = 13 hyp 12 Therefore, tanβ = 5 cosβ = Now use the identity tan2θ = 2tanθ 1- tan 2θ ⎛ 12 ⎞ 2⎜ ⎟ ⎝ 5 ⎠ tan2θ = 2 ⎛ 12 ⎞ 1- ⎜ ⎟ ⎝ 5 ⎠ 24 5 tan2θ = 144 125 24 5 tan2θ = 25 144 25 25 24 tan2θ = 5 119 − 25 24 25 tan2θ = ×5 119 120 tan2θ = 119 Trigonometry II Standards Test ANSWERS 16 www.math40s.com