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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