π ππππ½ + π ππππ½ General Solution cos ππ₯ = π 2ππ ± πΌ π₯= π sin ππ₯ = π ππ + −1 π πΌ π₯= π πβπππ πΌ = sin−1 π πβπππ πΌ = cos −1 π tan ππ₯ = π ππ + πΌ π₯= π πβπππ πΌ = tan−1 π πΉπππ(π − πΆ) π cos(π₯ − πΌ) = π (cos π₯ cos πΌ + sin π₯ sin πΌ) π cos(π₯ − πΌ) = π cos π₯ cos πΌ + π sin π₯ sin πΌ So, if we want to write an expression of the form a cos π₯ + π sin π₯ in the form π cos(π₯ − πΌ) we can do this by comparing π cos π₯ + π sin π₯ = π cos π₯ cos πΌ + π sin π₯ sin πΌ πΉπππ(π − πΆ) For both sides to be equal the terms in cos π₯ and sin π₯ must be equal. Therefore we have to equate the coefficients. π cos π₯ + π sin π₯ = π cos π₯ cos πΌ + π sin π₯ sin πΌ π cos π₯ = π cos π₯ cos πΌ π π ππ π₯ = π π ππ π₯ π ππ πΌ π = π cos πΌ π = π π ππ πΌ πΉπππ(π − πΆ) We can now find R and α in terms of π πππ π π2 + π 2 = π 2 cos 2 πΌ + π 2 sin2 πΌ π2 + π 2 = π 2 (cos 2 πΌ + sin2 πΌ) π2 + π 2 = π 2 π = π2 + π 2 πΉπππ(π − πΆ) We can now find R and α in terms of π πππ π Recall π = π cos πΌ π = π π ππ πΌ π π sin πΌ = π π cos πΌ π tan πΌ = π πΌ= tan−1 π π π cos π₯ + π sin π₯ can be written as π cos(π₯ − πΌ) where: π = πΌ= π2 + π 2 tan−1 π π π πΆ π Examples Express 2 cos π + sin π in the form π cos(π − πΌ) π€βπππ π > 0, 0° < πΌ < 90° 2 = π cos πΌ 1 = π sin πΌ π=2 π=1 π= π2 + π2 π= 12 + 22 π= 5 −1 π π tan−1 1 2 πΌ = tan πΌ= πΌ = 26.6° ∴ 2 cos π + sin π = 5 cos(π − 26.6°) Examples Express 3 cos π + 4 sin π in the form π cos(π − πΌ) π€βπππ π > 0, 0° < πΌ < 90° 3 = π cos πΌ 4 = π sin πΌ π=3 π=4 π= π2 π= 42 + 32 π=5 + π2 −1 π π tan−1 4 3 πΌ = tan πΌ= πΌ = 53.1° ∴ 3 cos π + 4 sin π = 5 cos(π − 53.1°) Examples Express 2 sin π − cos π in the form π sin(π − πΌ) π€βπππ π > 0, 0° < πΌ < 90° 2 sin π − cos π = π sin(π − πΌ) 2 sin π − cos π = π (sin π cos πΌ − cos π sin πΌ) 2 sin π − cos π = π sin π cos πΌ − π cos π sin πΌ 2 sin π = π sin π cos πΌ 2 = π cos πΌ π 2 sin2 πΌ + π 2 cos2 πΌ = 12 + 22 π 2 (sin2 πΌ + cos 2 πΌ) = 12 + 22 ππππππ: sin2 π₯ + cos 2 π₯ = 1 π 2 = 5 − cos π = −π πππ π π ππ πΌ 1 = π sin πΌ ∴ 2 sin π − πππ π = 5 π ππ(π − 26.6°) π = 5 1 πΌ = tan 2 πΌ = 26.6° −1 Examples Express 7 cos π₯ − 24 sin π₯ in the form π cos(π + πΌ) π€βπππ π > 0, 0° < πΌ < 90° Hence, solve the equation 7 cos π₯ − 24 sin π₯ = 4 for 0° < π₯ < 360° 7 = π cos πΌ 24 = π sin πΌ π=7 π = 24 π= 72 + 242 π = 25 24 πΌ= 7 πΌ = 73.7° tan−1 ∴ 7 cos π₯ − 24 sin π₯ = 25 cos(π₯ + 73.7°) Examples Express 7 cos π₯ − 24 sin π₯ in the form π cos(π + πΌ) π€βπππ π > 0, 0° < πΌ < 90° Hence, solve the equation 7 cos π₯ − 24 sin π₯ = 4 for 0° < π₯ < 360° 25 cos π₯ + 73.7° = 4 4 cos π₯ + 73.7° = 25 4 −1 π₯ + 73.7 = cos 25 π₯ + 73.7 = 80.8°, 279.2° π₯ = 7.1°, 205.5° Examples Express 2 cos π₯ + sin π₯ in the form π cos(π − πΌ) π€βπππ π > 0, 0° < πΌ < 90° Hence, solve the equation 2 cos π₯ + sin π₯ = 1 for 0° < π₯ < 360° 2 = π cos πΌ 1 = π sin πΌ π=2 π=1 π= 12 + 22 π= 5 1 πΌ= 2 πΌ = 26.6° tan−1 ∴ 2 cos π₯ + sin π₯ = 5 cos(π₯ − 26.6°) Examples Express 2 cos π₯ + sin π₯ in the form π cos(π − πΌ) π€βπππ π > 0, 0° < πΌ < 90° Hence, solve the equation 2 cos π₯ + sin π₯ = 1 for 0° < π₯ < 360° 5 πππ π₯ − 26.6° = 1 1 cos π₯ − 26.6° = 5 1 −1 π₯ − 26.6 = cos 5 π₯ − 26.6 = 63.4°, 296.6° π₯ = 90, 323.2°