# Order Trigonometric Functions Discussion

Order Trigonometric Functions Discussion
Order Trigonometric Functions Discussion
. As shown in this problem, fractions of a degree can be expressed in
minutes (′) and seconds (″). Recall that 1 minute = and 1 second =
392. 57.5692º = 57º (.5692)60′ = 57º 34.152′ = 57º 34′ (.152)60″ = 57º
34′ 9″
393. Using the 30º – 60º – 90º triangle, .
394. Using the 30º – 60º – 90º triangle, .
395. Using the 45º – 45º – 90º triangle, .
396. Using the unit circle, .
397. Using the 30º – 60º – 90º triangle, .
398. Using the unit circle, .
399. Using the unit circle, .
400. Using the 45º – 45º – 90º triangle, .
401. Using the 45º – 45º – 90º triangle, .
402. Using the unit circle, undefined.
403. The reference angle for 135º is 45º. The cosine is negative in quadrant II
so, .
404. The reference angle for 210º is 30º. The tangent is positive in quadrant III
so, .
405. Using the unit circle, .
406. Using the unit circle, .
407. Using the unit circle, .
408.
409.
410.
411.
412.
413. (a) phase shift: 2π units left; (b) vertical shift: 4 units up
414.
Order Trigonometric Functions Discussion
(a) phase shift: units right; (b) no vertical shift
415.
(a) phase shift: units left; (b) vertical shift: 1 unit down
416.
(a) phase shift: 3π units right; (b) no vertical shift
417.
(a) phase shift: units right; (b) vertical shift: units up
418. A = 3
419. A = 4
420. A =
421. A = π
422. A = 12
423. (a) period = ; (b) no phase shift; (c) no vertical shift
424.
(a) period = ; (b) phase shift: units right; (c) vertical shift: 5 units
down
425.
(a) period = = 1; (b) phase shift: units right; (c) vertical shift: 10 units
up
426.
(a) period = ; (b) phase shift: units left; (c) no vertical shift
427. y = cos(3x – 6) – 2 = cos(3(x – 2)) – 2
(a) period = ; (b) phase shift: 2 units right; (c) vertical shift: 2 units
down
Order Trigonometric Functions Discussion
428. The solutions are x = –2π, – π, 0, π, 2π because in the interval [–2π, 2π] the
graph of y = sin x intersects the x-axis at these values for x.
429. The solutions are because in the interval [–2π, 2π ] the
graph of y = cos x intersects the x-axis at these values for x.
430. ; amplitude = 3, period = , phase shift:
right, no vertical shift
431. y = –2cos(–3x – 6π) = –2cos(–3(x 2π)); refection about the x-axis,
amplitude = 2, period = , phase shift: 2π left, no vertical shift
432. , phase shift: 2 units right, no vertical shift
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