SECTION 7.5 Sum and Difference Formulas 515 Solution (a) Because α = = y r sin 4 5 and π α π < < 2 , let = y 4 and = r 5, and place α in quadrant II. The point ( ) ( ) = = P x y x , , 4 , < x 0, is on a circle of radius 5, + = x y 25. 2 2 See Figure 33. Then + = + = = − = =− x y x x x 25 16 25 25 16 9 3 2 2 2 2 Then α = = − x r cos 3 5 Alternatively, we can use α = sin 4 5 and the Pythagorean Identity α α + = sin cos 1 2 2 to find α cos . α α =−− =−− =− =− cos 1 sin 1 16 25 9 25 3 5 2 ↑ in quadrant II, cos 0 α α< (b) Because β = − = y r sin 2 5 and π β π < < 3 2 , let = − y 2 and = r 5, and place β in quadrant III.The point ( ) ( ) = = − P x y x , , 2, < x 0, is on a circle of radius 5, + = x y 5. 2 2 See Figure 34. Then + = + = = = − x y x x x 5 4 5 1 1 2 2 2 2 Then β = = − = − x r cos 1 5 5 5 Alternatively, use β = − sin 2 5 5 and the Pythagorean Identity β β + = sin cos 1 2 2 to find β cos . β β =−− =−−=− =− cos 1 sin 1 4 5 1 5 5 5 2 (c) Use the results found in parts (a) and (b) and the Sum Formula for cosine. α β α β α β ( ) + = − =− − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ − − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ = cos cos cos sin sin 3 5 5 5 4 5 2 5 5 11 5 25 (d) Use the Sum Formula for sine. α β α β α β ( ) ( ) + = + = − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ + − − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ = sin sin cos cos sin 4 5 5 5 3 5 2 5 5 2 5 25 Figure 34 β π β π = − < < sin 2 5 , 3 2 b y x x2 1 y2 5 5 (x, 22) 2 5 5 2 5 5 5 Figure 33 α π α π = < < sin 4 5 , 2 a y x x2 1 y2 5 25 (x, 4) 25 25 5 5 5 = y 4 <x 0 =− y 2 <x 0 Now Work PROBLEMS 35(a), (b), AND (c)
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