Amplitude, phase and vertical shift.

Dear experts,

I am studying trigonometry functions.

In trigonometry functions, we have y=sin-theta and y=a.sin (3.theta + 30 degrees) + 60 degrees

I understand that when y=sinx we use the unit circle with radius of 1unit to explain radian/amplitude.

I am trying to under the following concepts of amplitude, phase shift and vertical shift. I understand the concept of period for trigonometric functions.

Now is there a sequence of videos or notes which will help me understand these concepts. Not just the calculations but also the implications of unit circle and on the trigonometry functions. My struggle is not just to crunch numbers but also to understand the concepts how the numbers are impacting the trigonometric functions.

Kindly guide.
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"I understand that when y=sinx": make sure you understand where cos and tan come from with the unit circle.  That will help with some the relationships between them.

I wouldn't really put amplitude and vertical shift in the realm of trig, though phase shift could apply.  You can model phase shift on the unit circle (add or subtract the appropriate angle), but I'm not sure that helps.  

You really need to be looking at a graph with amplitude vs. angle (or time) as the line in the unit circle rotates.  A sine or cosine wave will have the smooth, symmetrical up and down shape as you move along the X axis.  The amplitude is the vertical height of the wave (commonly measured from peak-to-peak).  Phase shift would be the wave sliding to the right or left along the X axis.  Vertical shift would be the wave sliding up or down along the Y axis.  "Standard" sine or cosine waves are symmetrical around the X axis.  Once you have vertical shift, it's no longer symmetrical.

This will give you some information that may make it understandable: .
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>> a.sin (3.theta + 30 degrees) + 60 degrees

The argument to the sin function is degrees or radians as you have shown. You are not supposed to add degrees to a sin function (at least not to my understanding of usual applications of trig problems).
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I concur with phoffric.  The result of a sin function is a unitless number.  If you add 60 degrees to that, you get a unitless number + 60 degrees, which is very odd.  It's like having 6 apples + 1 as opposed to 6 apples + 1 apple.
Concurring with phoffic this equation is meaningless "y=a.sin (3.theta + 30 degrees) + 60 degrees".

As I explained in your other question sine returns a value which is the ratio of two sides of a triangle and the angle between them. Since this is a ratio, it can be of a triangle of ANY size, not just a triangle based on a unit circle.

Since sine, the ratio of the ordinate to the hypotenuse, and cosine, the ratio of the abscissa to the hypotenuse are related to a triangle and ALWAYS are ratios to the hypotenuse, then by Pythagoras - the square of the two sides equals the square of the hypotenuse, NEITHER sine nor cosine can exceed 1. Therefore in y=a*sin(x) a is the amplitude.

Since the triangle is a right angled triangle the sin of the other angle will be the ratio of the other side to the hypotenuse. Since this side is the abscissa it is therefore the cosine of the angle. This angle is 90°-our angle and we see immediately that the sin(180° - 90° - a)  = cos(a) . Thus cosines start 90° after sines.

For a phase shift we must compare two "waves" of the SAME frequency (or period).  The difference along the abscissa of two points with ther same values on each curve is called a phase shift (in relation to ONE of the curves). Clearly the cosine to the sine has a 90° phase shift, from sin(90°-a)=cos(a), and hence in sin(x+a) the "a" represents the phase shift w.r.t sin(x().

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ExcellearnerAuthor Commented:
I missed to award points to CompProbSolv. Can you please reopen the question.
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