Question

3.1.1 use your calculator to complete the following table:

θ	30°	0°	60°	150°	240°	330°
p(θ)=sin(θ + 30°)

3.1.2 the graph of f(θ)=sin θ, θ ∈ 0°, 360°, is drawn on the set of axes below. use the table above to sketch, on the same set of axes, the graph of p(θ)=sin(θ + 30°), θ ∈ -30°, 330°.
compare the graph you have drawn to the graph of f(θ)=sin θ, note the similarities and differences and answer the questions that follow.
3.1.3 what is the period of p(θ)=sin(θ + 30°)?
3.1.4 what is the range of p(θ)=sin(θ + 30°)?
3.1.5 what is the amplitude of p(θ)=sin(θ + 30°)?
3.1.6 explain in your own words what effect the +30° has on changing the graph of f into graph p.
3.1.7 why do you think the table was started at 30°?

Explanation:

Step1: Calculate p(30°)

$\theta=30^\circ$, $\theta+30^\circ=60^\circ$, $\sin(60^\circ)\approx0.866$

Step2: Calculate p(0°)

$\theta=0^\circ$, $\theta+30^\circ=30^\circ$, $\sin(30^\circ)=0.5$

Step3: Calculate p(60°)

$\theta=60^\circ$, $\theta+30^\circ=90^\circ$, $\sin(90^\circ)=1$

Step4: Calculate p(150°)

$\theta=150^\circ$, $\theta+30^\circ=180^\circ$, $\sin(180^\circ)=0$

Step5: Calculate p(240°)

$\theta=240^\circ$, $\theta+30^\circ=270^\circ$, $\sin(270^\circ)=-1$

Step6: Calculate p(330°)

$\theta=330^\circ$, $\theta+30^\circ=360^\circ$, $\sin(360^\circ)=0$

Step7: Determine period of p(θ)

Sine period is 360°, phase shift doesn't affect.

Step8: Determine range of p(θ)

Sine range is [-1,1], unchanged by phase shift.

Step9: Determine amplitude of p(θ)

Amplitude of sin(θ+c) is 1 (no coefficient).

Step10: Effect of +30° on graph

Shifts f(θ) left by 30°.

Step11: Reason table starts at 30°

p(θ) domain [-30°,330°], 30° is key point.

Answer:

3.1.1 Table values (from left to right): 0.866, 0.5, 1, 0, -1, 0
3.1.3 360°
3.1.4 [-1, 1]
3.1.5 1
3.1.6 It shifts the graph of f(θ) to the left by 30°.
3.1.7 Because the domain of p(θ) is [-30°, 330°], and starting at 30° helps capture key points of the function.