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# Trigonometry: FAQ

## What is the unit circle and why is it important in trigonometry?

The unit circle is a circle with a radius of $1$ that is centered at the origin on a coordinate plane. It's important in trigonometry because it allows us to define the sine and cosine functions in terms of the $x$- and $y$-coordinates of a point moving around the circle.

## What are radians and why do we use them in trigonometry?

Radians are a unit of measurement for angles.
One radian is the angle measure that we turn to travel one radius length around the circumference of a circle.
We often use radians in trigonometry because they make working with trigonometric functions easier.

## What is the Pythagorean identity and why is it important?

The Pythagorean identity is ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$. It comes from the Pythagorean theorem and is important in trigonometry because it can help us solve for the value of one trigonometric function if we know the other.

## What do amplitude, midline, and period mean when we're talking about sinusoidal graphs?

The amplitude of a sinusoidal graph is the distance from the midline to the highest or lowest point on the graph. The midline is the horizontal line that the graph oscillates around, and the period is the horizontal distance it takes for the graph to complete one full cycle.

## Why do we want to know how to transform sinusoidal graphs?

By understanding how to transform sinusoidal graphs, we can graph a wider variety of sinusoidal functions. For example, we can change the amplitude, midline, or period to match a given equation.

## Where are trigonometric functions used in the real world?

Trigonometric functions are used in many real-world applications. For example, engineers use them to design bridges, and physicists use them to model periodic phenomena such as waves or vibrations.

## Want to join the conversation?

• Did the modeling w/ sinusoidal functions (phase shift) unit; how to know when to do sin or cos since either can happen w/ a phase shift? Instructions don't specify which/my answer satisfies requirements, but it isn't counted as correct.
• As we know, when looking at a unit circle, cosine's value will reach 1 when x=0 radians, whereas sine's value will reach 1 when x=pi/2 radians. In a scenrio where the graph reaches its maximum point when x=0, you know that it will be a graph of cosine. However, if you have a scenario where you reach the midline (midpoint) at a x-value of 0, you know you have a graph of sine. Essentially, you want to look at where you reach your maximum or midline when analyzing what type of a graph to use. Now, if you are given a problem with a phase shift, you want to look at the maximum and minium and shift it with relevance to what format (sine or cousin) your graph is in. I would suggest looking at one of the videos that Sal posted. He really thoroughly explains how to know what format to use.
• I really do not understand the substitution in phase shifting a function. I've tried it in every way I can imagine solving it... for example: −1.5cos(2π/248(22+11))+5.9 , solved in several different ways, always evaluated to approx. 4.4002.... I got those questions wrong every single time. what am I doing wrong here? what am I missing??
• could you show me your work? what are the steps you take to get this answer? Also, what is the problem? This is how I would solve it:

1)(33π/248) ≈ 0.836069...

2) cos(0.836069) ≈ 0.6703848... MAKE SURE TO USE RADIAN MODE FOR THIS STEP. The khan academy calculator(and most others) have a deg and rad mode for trig functions.

3) -1.5*0.670384... ≈ -1.0055..

4) -1.0055...+5.9 = 4.894422...

Hope this helped! Next time show your work as well :)
• I am struggling to understand one of the problems I encountered where I was asked WHEN the function would hit it's maximum.

There was a function given,

h(t) = 5 - 2 * sin(2pi (t + 1) / 7)

I understand that the function is negative so the it's a sine reflection.

I think I also understand that the function ignoring the +1 would reach it's maximum at t=5.25 (i.e. 3/4 * 7)

what I don't understand is why when considering the +1 the max takes place at t=6.25 instead of t=4.25?

Shouldn't a +1 SHIFT the whole function to the LEFT - causing the MAX to be encountered sooner by 1?

Thank you.
• unless you are reading the problem wrong(which happens to the best of us), then you are right and the answer is wrong. It should be 4.25. Maybe it's -1 and not +1?
• On one of the practice quizzes the answer in part only was "-357cos(...". I came up with the same answer except I had as positive instead of a negative, i.e. "357cos(...". I tried them both on Desmos and the -357 did not appear correct to me. Did I miss something? I admit some of these problems have been a challenge.
• the negative flips the graph across the x-axis. This could mean that in your problem, the starting value needed to be the minimum point(this might not be the case since you did not provide the question)
• What's an easy way to remember your formulas for period?
• The answer to the practice quiz about temperature in Guangzhou, China is confusing. How do you read the given graph? The x component is what I don't understand. It goes from 0 to beyond 2.5. I suppose that at 2.5 it means 2.5 years after 1/1/2015. But how does it model daily lows at all. Daily lows would in reality fluctuate around the cos function, and the cos function is modeling the average over the year. On any one day it would give you an average temperature for that day. Let's say the date is July 26 as mentioned in the question. How do you get a daily low out of that? It is the same with any and every other day on the graph. You don't have that kind of accuracy. If the question asked for the daily average temperature t years after 1/1/2015 it would make sense. Why does it ask for daily lows? I don't get it.
• What's an easy way to remember your formulas for period?
• Is a sine function a phase shift of a cosine function and vice versa?
(1 vote)
• Yes. sin(x+π/2)=cos(x).
• How do we determine the amplitude sign in this equation? The analysis has not been understood.
• The sign of the amplitude in a sinusoidal equation (i.e. whether it is negative or positive) is a matter of whether your graphed function is horizontally inverted. This refers to your function being flipped in the x-(and y-)axis. Although, your amplitude is always the same regardless of its positive/negative expression.
To understand horizontal inversions, you first need to know what a regular sinusoid (with a positively expressed amplitude) looks like. For example, a regular function of y=cos(x) shows the y-intercept as a maximum point with coordinates (0,+1). However, in the function y=-cos(x), the y-intercept is flipped and is now a minimum point with coordinates (0,-1).

Regarding the sine function, the regular graph of y=sin(x) shows the y-intercept as the midline with coordinates (0,0), with the closest maximum point on its right the closest minimum point on its left. However, in y=-sin(x), the closest maximum point is now on the left of the y-intercept and the closest minimum point is on its right. You should toggle with these functions to actually visualise what I mean.

With this said, the sign in front of the coefficient for amplitude (a) depends on whether your function has been horizontally inverted. So, to actually determine whether it's positive or negative, you must look at the behaviour of your graph and compare it to a regular sinusoid with the sine/cosine function. If it's flipped, your amplitude is expressed as negative (-); if it's not flipped, your amplitude is expressed as positive (+) in the equation.

Hope this helps.
(1 vote)
• 727+22sin(2𝜋/365 (𝑥−80.75))
This is one of the functions that I got from a question.
My graphing calculator wouldn't take it because of the decimal! Why?
(1 vote)
• You may have to put 2*pi instead of 2pi on the calculator, at least that is a common mistake I have.
(1 vote)