Question

1. Use the graph of y=sin 2x to find the value of sin 2x for x=pi/4 radians.

A. -1
B. 0
C. 0.5
D. 1

2. What is an equation for a sine curve with amplitude 2, and period 4pi radians.

A. y = 0.5sin4x
B. y = 4sin (x/2)
C. y = 2sin2x
D. y = 2sin (x/2)

3. Sketch one cycle of the cosine function.
y = 2cos2x

C. graph*

4. Find the period, range, and amplitude of the cosine function.
y = -3cos4x

A. period 3pi; range -3 <= y <= 3; amplitude = 1/2
B. period 1/2; range -3 <= y <= 3; amplitude = -3
C. period 1/2pi; range -3 <= y <= 3; amplitude = 3
D. period 3; range -3 <= y <= 3; amplitude = 1/2

Answer 1

1)
$y = \sin(2x) \\ y = \sin(2( \frac{\pi}{4} ) ) \\ y = \sin( \frac{\pi}{2} ) \\ y = 1$
D is the answer for 1

2)
$y = 2\sin( \frac{x}{2} )$
3) Picture*** ( Sorry I'm on the bus).

4) Choice C.

Answer 2

Answer:

Step-by-step explanation:

Ans 1.)

The graph of $y=sin2x$ in fig-1

When put $x=\frac{pi}{4}$ in $y=sin2x$ so, we get

$y=sin2(\frac{pi}{4})$

$y=sin(\frac{pi}{2})$

We can see the value of $y=sin(\frac{pi}{2})$ in figure-1,

the value of $y=sin(\frac{pi}{2})$ is 0.

Hence the correct option is (B).

Ans 2.)

The general equation is $y=Asin(B(x+C))+D$

Where Amplitude=A,

$\text{period}=\frac{2\pi}{B}$

Phase shift =C

and Vertical shift=D

So, if Amplitude is 2 then A=2 and period is 4pi radian then $\frac{2pi}{4pi}$,

We get equation as;

$y=2sin(\frac{x}{2})$

Hence correct option is (D).

Ans 3.)

one cycle of  cosine function is shown in figure-2

Ans 4.)

The general equation is $y=Acos(B(x+C))+D$

Where Amplitude=A,

$\text{period}=\frac{2\pi}{B}$

Phase shift =C,

Vertical shift=D and range is $-A\leq y\leq A$.

In cosine function, $y=-3cos4x$;

Period is $\text{B}=\frac{1}{2pi}$, Range is $-3\leq y\leq 3$ and Amplitude is 3.

Hence, the correct option is (C).