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( (\pi)/(4) ) ) \\ y = \sin( (\pi)/(2) ) \\ y = 1`
D is the answer for 1

2)

`y = 2\sin( (x)/(2) )`
3) Picture*** ( Sorry I'm on the bus).

4) Choice C.

Answer 2

Explanation:

Ans 1.)

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

When put
`x=(pi)/(4)` in
`y=sin2x` so, we get

`y=sin2((pi)/(4))`

`y=sin((pi)/(2))`

We can see the value of
`y=sin((pi)/(2))` in figure-1,

the value of
`y=sin((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}=(2\pi)/(B)`

Phase shift =C

and Vertical shift=D

So, if Amplitude is 2 then A=2 and period is 4pi radian then
`(2pi)/(4pi)`,

We get equation as;

`y=2sin((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}=(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}=(1)/(2pi)`, Range is
`-3\leq y\leq 3` and Amplitude is 3.

Hence, the correct option is (C).