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

A. y=0.5sin 4 theta
B. y=4sin 0/2
C. y=2sin2 0
D.y=2sin0/2

Find the period and amplitude of the function.

Y=5 cos 2 0theta

A. Period: pi radians; Amplitude:2.5
B. Period: pi radians: Amplitude:5
C. Period: pi radians: Amplitude:2.5
D. Period: pi radians; Amplitude:5

Tan 2pi/3

A. sqrt3
B. 1/sqrt3
C. -sqrt3
D. 1

Tan 5pi/6

A. sqrt3
B.-sqrt3/3
C.-sqrt3
D.1

2 Answers

5 votes

Answer:

B.)

Explanation:

Got it right on Edge

User Justinrixx
by
5.1k points
1 vote
1. a sine curve with amplitude 2, and period 4pi radians

the general equation of the sine curve ⇒⇒ y = a sin (nθ)
where: a is the amplitude and n = 2π/perid
amplitude 2, and period 4pi radians

∴ y = 2 sin (θ/2)

The correct answer is option D. y = 2 sin (θ/2)
===========================================

2.The period and amplitude of the function ⇒⇒ y = 5 cos 2θ

comparing with y = a cos nθ

where : a is the amplitude and n = 2π/period
amplitude = 5 , period = 2π/n = 2π/2 = π


The correct answer is option B. Period: pi radians: Amplitude:5

============================================================
3. tan (2π/3) = tan 120° = -√3
120° lie in the second quadrant and its reference angle = 180° - 120° = 60°
tan function in the second quadrant is negative
∴ tan 120° = - tan 60 = -√3

The correct answer is C. -sqrt3

=====================================================
4. Tan 5π/6 = tan 150° = -(√3)/3

150° lies in the second quadrant and its reference angle = 180° - 150° = 30°
tan function in the second quadrant is negative
∴ tan 150° = - tan 30 = -(√3)/3

The correct answer is B.-sqrt3/3
User Vishal Shah
by
5.3k points
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