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

tan-x= -tanx =-sinx/cosx

tan-x)(sinx)=-sin^2(x)/cosx= (cos^2(x)-1)/cosx= cosx-secx

sec-x =secx =1/cosx

secxcos^2(x)= (1/cosx)(cosx)(cosx)= cosx

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

=cosx -secx - cosx

=secx

2a) period =2pi/B= 2pi/5

phase shift =C = -2

g(x) = AsecB(x +C) + D

for g(x) =-2sec5(x-2) + 3

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

parent function has period of pi

but with pi/12 as a coefficient of x the period is compressed by a factor of 12/pi. 12/pi times pi = 12

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

period = pi/B = pi/pi/12 = 12

phase shift = 2

3) a=5 c =sqr74

sqr74 - 5 < b < sqr74 +5

0 < angle B < 180

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

2cos(pix/5) = sqr3

cos(pix/5) = sqr3/2

pix/5 = pi/6 or -pi/6

x = (5/pi)(pi/6) or (5/pi)(-pi/6)

x = 5/6 or -5/6

5)a) sin^-1(-sqr2/2)

= -pi/4

5)b) cos^-1(-1/2)

=2pi/3