The sine curve y = a sin(k(x - b)) has amplitude amplitude period , period , and horizontal shift and horizontal shift = 5 sin(7(x-7)) h has The sine curve y = 5:

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A: Here we use the general equation of tangent function that is y=a tan(bx) and the period =πb

Q: cos-1 (cosx)= x       where..............

A: Given, cos-1cos x=x

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A: The given function is  y=2sin12x-π4               ⋯1 Consider the standard sin function with…

Q: How can the graph of y = sin-1 x be obtained from the graph of the restricted sine function?

A: given: y=sin-1x

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A:

Q: es of the five quarter points of y=sinx to determine the corresponding quarter points on the graph…

A: Introduction:

Q: CSc : 1/ sin :: sec :

A: Evaluate csc:1sin::sec:?

Q: t-a) Scin 3ada

A: L= ∫0t(t-a) sin3a da   =∫0tt sin3a da- ∫0ta sin3a da   =-t cos 3a3t0- a∫0t sin3a da+ ∫0t dada…

Q: hat is the period, amplitude, phase shift of y=-4sin2(x-45

A: calculation of above requirement are as follows

Q: 15

A: We have to find equation

Q: y= Scos ( - ) (G-)) Amplitude: 3 Period 47 Start (Horizontal Shift):

A: given  y=3cos12x-π4

Q: y=$sin(4x-

A:

Q: Graph y= 2 cos(1/2x)+5 on the interval [0,2pi]. Explain amplitude, period and midline.

A: Note that 1 radian =57.27 deg approx. So you need to check the part of graph from 0 to 6.28 approx.…

Q: Summarize how the derivatives of inverse trigonometric functions are related to the derivatives of…

A: dsin-1xdx=11-x2dcos-1xdx=-11-x2dtan-1xdx=11+x2dcot-1xdx=-11+x2dsec-1xdx=1xx2-1dcsc-1xdx=-1xx2-1