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Name ________________________________________ Date __________________ Class__________________ Reteach LESSON 11-5 Double-Angle and Half-Angle Identities There are special double-angle formulas for sin2θ, cos2θ, and tan2θ. Double-Angle Identities The double angle is twiceθ, or 2θ. sin2θ = 2sinθ cosθ cos2θ = cos2θ − sin2θ There are three equivalent forms of the identity for cos2θ. If you know one form, you can use sin2θ + cos2θ = 1 to derive the other forms. cos2θ = 2cos2θ − 1 cos2θ = 1 − 2sin2θ tan 2θ = 2 tanθ 1 − tan2 θ Find sin2θ and cos2θ if cos θ = − Step 1 Find sinθ. sin2θ = 1 – cos2θ ⎛ 2⎞ sin2 θ = 1 − ⎜ − ⎟ ⎝ 3⎠ 5 2 sin θ = 9 sin θ = − 2 Pythagorean identity 2 Substitute cos θ = − . 3 Notice the interval in which θ lies. Cosine and sine are both negative in Quadrant III. Evaluate. 5 3 Use the negative value since θ lies in Quadrant III. Solve for sinθ. Step 2 Find sin2θ. sin2θ = 2sinθ cosθ ⎛ 5 ⎞⎛ 2 ⎞ − sin2θ = 2 ⎜ − ⎜ 3 ⎟⎟ ⎜⎝ 3 ⎟⎠ ⎝ ⎠ sin2θ = 2 and 180° < θ < 270°. 3 4 5 9 Write the formula. Substitute values for sinθ and cosθ. Simplify. Step 3 Find cos2θ. cos2θ = 2cos2θ – 1 2 ⎛ 2⎞ cos 2θ = 2 ⎜ − ⎟ − 1 ⎝ 3⎠ 8 1 cos 2θ = − 1 = − 9 9 Choose a formula. Substitute the value for cosθ. Simplify. Evaluate each expression if cos θ = 1. sinθ Use sin2θ = 1 − cos2θ. 1 and 270° < θ < 360°. 4 2. cos2θ Use cos2θ = 1 − 2sin2θ. _________________________________________ 3. tanθ ________________________________________ 4. tan2θ sin θ . Use tan θ = cos θ Use tan 2θ = _________________________________________ 2 tanθ . 1 − tan2 θ ________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 11-38 Holt McDougal Algebra 2 Name ________________________________________ Date __________________ Class__________________ LESSON 11-5 Reteach Double-Angle and Half-Angle Identities (continued) You can use half-angle identities to find the exact value of some trigonometric expressions. Half�Angle Identities 1 − cos θ θ θ ⎛θ ⎞ sin ⎜ ⎟ = ± : (+) for in Quadrants I, II; (−) for in Quadrants III, IV 2 2 2 2 ⎝ ⎠ 1 + cos θ θ θ ⎛θ ⎞ cos ⎜ ⎟ = ± : (+) for in Quadrants I, IV; (−) for in Quadrants II, III 2 2 2 ⎝2⎠ 1 − cosθ θ θ ⎛θ ⎞ tan ⎜ ⎟ = ± : (+) for in Quadrants I, III; (−) for in Quadrants II, IV 2 2 1 + cos θ ⎝2⎠ Think: 225° has a reference angle of 45°. Find the exact value of cos 112.5°. 225° 2 1 + cos θ =− 2 1 + cos 225° =− 2 cos112.5° = cos ⎛ 2⎞ 1 + ⎜⎜ − ⎟ 2 ⎟⎠ ⎝ =− 2 ⎛ 2 − 2 ⎞⎛ 1⎞ =− ⎜ ⎜ 2 ⎟⎟ ⎜⎝ 2 ⎟⎠ ⎝ ⎠ =− Find the angle: θ = 225°. Choose the half-angle identity. Substitute θ = 225°. Evaluate. Simplify. 112.5° is in QII where cosine is negative. cos225° = −cos45° 2 =− 2 2− 2 2− 2 =− 4 2 Use the half-angle identity to find the exact value of each expression. 315° 30° 6. sin15° = sin 5. tan 157.5° = tan 2 2 1 − cos 315° ⎛ 315° ⎞ tan ⎜ =− ⎟ 1 + cos 315° ⎝ 2 ⎠ ⎛ 30° ⎞ sin ⎜ ⎟ = _____________________ ⎝ 2 ⎠ _________________________________________ ________________________________________ _________________________________________ ________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 11-39 Holt McDougal Algebra 2 Reteach 1. − 15 4 3. − 15 2. − 7 8 4. 15 7 5. tan 2 2− 2 2 ;− 2 ; 5. − 2 2+ 2 1+ 2 2 6. 7. tan sin A 1 − cos A sin A 1. a. sin 2θ = 2 sinθ cosθ b. d (θ ) = c. 35 = 3v 0 2 sin 2θ 32 3(20)2 sin 2θ 32 d. sin2θ = 0.933 Challenge e. 34.5° sin 2 A cos 2 A 2. a. 2 sin A cos A b. tan 2 A = cos2 A − sin2 A Angle of Elevation (°) Horizontal Distance (yd) 15 18.75 25 28.73 35 35.24 45 37.50 55 35.24 2 sin A cos A cos2 A c. tan 2 A = 2 cos A sin2 A − cos2 A cos2 A 2 tan A 1 − tan2 A 2. tan2A = tan(A + A) = tan A + tan A 2 tan A = 1 − tan A tan A 1 − tan2 A 3. a. tan 2 A = = Problem Solving 2− 3 2− 3 ; 4 2 d. tan 2 A = 2 2 3 1− 2 ; 2 1. a. tan 2 A = A 1 − cos A 1 − cos A = ⋅ 2 1 + cos A 1 − cos A (1 − cos A ) 2+ 2 2+ 2 2− 2 2− 2 2− 2 2 ⋅ = ⋅ = 2 −1 2+ 2 2− 2 2 2 b. 2− 2 1 − cos 30° ; 2 2− 2 6. a. tan 22.5° = 1− − A 1 − cos A =± 2 1 + cos A b. 45° 3. 26.5° 4. Possible answer: For this initial velocity and target distance, sin2θ ≈ 1.32, which is outside the possible range of values for the sine of an angle; so he is incorrect. 24 ≈ 3.43 7 b. m∠A = 127°, tan254° ≈ 3.49 5. C 1 − tan2 A 4. a. 2 cot 2 A = tan A b. cot A − tan A = 6. G 1 − tan2 A tan A Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. A54 Holt McDougal Algebra 2