Periodic Functions

Question

1. Simplify tan(−𝑥) sin(𝑥) − sec(−𝑥) 𝑐𝑜𝑠2 (𝑥) into a single trig function. (Show all algebra work)
2. Find the period and horizontal shift for the following functions.
3. 𝑔(𝑥) = −2 sec(5𝑥 − 10) + 3
4. 𝑓(𝑥) = 3 cot ( 𝜋/12 𝑥 + 𝜋/6 ) − 1
5. Let 𝑎 = 5 and 𝑐 = √74. Solve for the unknown side and angle in degrees. Show all work and round to 3 decimal places
6. Solve 2 cos ( 𝜋/5 𝑥) − √3 = 0. For all solutions on two cycles. Show all algebra work. Provide an exact solution in radians(no decimals)
7. Evaluate the following inverse trig expressions, giving the exact answer in radians. (No decimals)
8. 𝒔𝒊𝒏^−𝟏 (− √𝟐/𝟐 )
9. 𝒄𝒐𝒔^−𝟏 (− 𝟏/𝟐 )

Answer 1

1) tan(-x)sin(x) -sec(-x)cos^2(x)

cos(x) and sec(x) are even functions

sin(x) is an odd function, sec(x)=1/cos(x), tan(x) = sin(x)/cos(x)

tan(-x)sin(x) - sec(-x)cos^2(x)

= sin(-x)/cos(-x)]sin(x) - (1/cos(-x))cos^2(x)

= (-sin(x)/cos(x)]sin(x)+(1/cos(x))(cos^2(x)

= -sin^2(x)/cos(x) +cos^2(x)/cos(x)

= (cos^2(x) +sin^2(x))/cos(x)

= 1/cos(x)

= sec(x)

2a) -2sec[2(x-2)]+3

AsecB(x+C)+D

B= 2 = 2pi/period, period = 2pi/B = 2pi/2 = pi

period = pi

horizontal phase shift= C = 2

2b) 3cot(pix/12 +pi/6) -1

= 3cot(pi/12)(x+2pi) -1

period = 2pi/B = 2pi/(pi/12) = 2pi(12/pi) = 24

phase shift = 2pi

3) a=5, c=sqr74, 3rd side

b = sqr(a^2 +c^2 -2abCosB)

a=4, c =sqr74

|sqr74 -5| < b < sqr74 +5

8.6- 5< b < 8.6+5

3.6 < b < 13.6

b^2 = (25+74 -2(5(sqr74)CosB

b^2 = 99-10sqr74CosB

CosB = (b^2-99)/10sqr74

Angle B = Cos^-1[(b^2-99)/10sqr74]

4) 2cos(pix/5) -sqr3 = 0

cospix/5 = sqr3/2

pix/5 = +/-30 degrees = +/-pi/6 +/-npi/3 where n= any integer

5) sin^-1(-sqr2/2) = -45 degrees = -pi/4 radians

domain is from -pi/2 to pi/2

cos^-1(-1/2) = 2pi/3 radians = 120 degrees

domain is 0 to pi