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Published bySeamus Killingbeck Modified over 3 years ago

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Example Express -8sinx° + 15cosx° in the form ksin(x + )° ********* a = -8 & b = 15 kcos°<0 Q2 or Q3 kcos° = -8 and ksin° = 15 ksin°>0 Q1 or Q2 (kcos°)2 + (ksin°)2 = (-8) k2 = k2 = 289 k = 17 tan° = b/a = 15/-8 in Q2 tan-1(15/8) = 61.9° Q2: = 180 – 61.9 = 118.1° So -8sinx° + 15cosx° = 17sin(x )°

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Example (radians) Express 3/2sin - 1/2cos in the form ksin( + ) ********* a = 3/2 & b = -1/2 kcos>0 Q1 or Q4 kcos = 3/2 and ksin = -1/2 ksin<0 Q3 or Q4 (kcos)2 + (ksin)2 = (3/2 )2 + (-1/2)2 k2 = 3/4 + 1/4 k2 = 1 k = 1 in Q4 tan = b/a = (-1/2)(3/2) = -1/2 X ( 2 /3) = -1/3 tan-1 (1/3)= 30° = /6 Q4: = 2 - /6 = 11/6 So 3/2sin - 1/2cos = sin( + 11/6)

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VARIATIONS The usual formats are acosx° + bsinx° = kcos(x - )° asinx° + bcosx° = ksin(x + )° The variations kcos(x + )° and ksin(x - )° are easily obtained by considering reverse rotations from the X-axis. NB: -200° = +160° while ° = -240° So sin(x + 225)° = sin(x – 135)° and cos( - 7/5) = cos( + 3/5)

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Example Express 12sinx° - 5cosx° in the form ksin(x - )° ********* a = 12 & b = -5 kcos° = 12 and ksin° = -5 kcos°>0 Q1 or Q4 (kcos°)2 + (ksin°)2 = (-5)2 ksin°<0 Q3 or Q4 k2 = k2 = 169 k = 13 tan° = b/a = -5/12 in Q4 tan-1(5/12) = 22.6° Q4: = 360 – 22.6 = 337.4° So 12sinx° - 5cosx° = 13sin(x )° = 13sin(x – 22.6)°

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2 step problems 5) Solve 0.5Cos(x) + 3 = 2.6 1) Solve 4Sin(x) = 2.6 2) Solve Cos(x) + 3 = 3.28 3) Solve 2Tan(x) + 2 = 5.34 4) Solve 2 + Sin(x) = 1.85 180.

2 step problems 5) Solve 0.5Cos(x) + 3 = 2.6 1) Solve 4Sin(x) = 2.6 2) Solve Cos(x) + 3 = 3.28 3) Solve 2Tan(x) + 2 = 5.34 4) Solve 2 + Sin(x) = 1.85 180.

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