How To Graph Trigonometric Functions | Trigonometry
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- čas přidán 13. 04. 2021
- This trigonometry video tutorial explains how to graph sine and cosine functions using transformations, horizontal shifts / phase shifts, vertical shifts, amplitude, and the period of the sinusoidal function. This video contains many examples and practice problems on graphing trigonometric functions for you to master this topic.
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The full version explains how to graph secant, cosecant, tangent and cotangent functions.
Full 1 Hour 25 Minute Video: www.patreon.com/MathScienceTutor
Direct Link to The Full Video: bit.ly/38wtIWU
But why change it to 4 pi over 2 instead of leaving it as 2 pi in the second last question y=sin( x - pi/2)
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MR. Organic Chemistry Tutor, thank you for showing high school and college students How to Graph Trigonometric Functions the correct way. You are the first instructor that showed me How to Graph Trigonometric Functions the correct way.
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First few mins in and I already understood a bit better thanks. My instructor in trig has been doing it for awhile and leaves vital info out during his lecture vids.
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I don't get it at the level of 15:15 because of the graph extending to 4pii and why is the negative y-axis is upto -3 .
I have my term final tomorrow and part of it dealt with cosine and sine functions. You are a life saver, I didn't understand this until just now
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For anyone wondering how he figures out the tick mark values on the x-axis: there are 5 critical points in the sin/cos cycle. Look at a one cycle of a cos/sin graph, and there will be 5 values: 2 values dictating the upper/lower bounds (y-axis values), and 3 values that will cross the x-axis (zeros). In order to find those critical point values (if you're dealing with a phase-shift), you find the value of the phase shift (0=Bx+c, solving for x), and then you add pi/2 recursively.
For example, for y=sin(x+pi/3), the graph will start at x=-pi/3. Critical point #1 is -pi/3. To find critical point #2, -pi/3 + pi/2 = pi/6. To find critical point #3, pi/6 + pi/2 = 2pi/3. And so on until you have 5 values. You can use a calculator to add everything to make it easier. if you're still having trouble, google LibreText Mathematics section 2.4 on Phase shift.
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There's nothing "video" here. The half-wit has entirely missed the opportunity to use a visual medium to show why these trigonometric ratios take the shape they do -- because radii stay th same while the other sides of the triangles involved vary.
Just a horrid, stupid, ignorant waste.
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To Sum Up: For equations y=a sin(bx) and y=a cos(bx), |a| is the highest y value point on the graph and -|a| is the lowest y value point on the graph. |b| represents the frequency of the graph or 2pi/b
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Thank you so much! You helped heaps, for some reason I don’t understand anything my teacher is saying ;(
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Hi,just wanna enquire act y do we put pie values and not actual real values such as 90,180,360 and so on🤔
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How did you get 3pie/2? 19:07
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What's the period of the function y= sin x + cosx???
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Thanks Mr, but that function of y=2cos(x)-1 the period is in 360 degrees, or am I confused.
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Hello incase you need help in online classes,assignments and tasks ,I'm here to help you out.
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you can also think of B as the number of cycles within a length of 2π
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Why is it so easy for me to understand you and so hard for me to understand my teachers?
at 17:45 isn’t the general form of sine function is asinx(bx-c)+d?
No it is + c
how do you know how many key points the graph are that's where I'm having a hard time
At 21:50 the number in between 5 and 1 should be 3 on the y-axis of the graph.
There is 2 at the middle. Its just cause the amplitude is 2 he shifted the graph 2 units up and down from 3 which will bring us to the numbers 5 and 1 respectively.
The equation is y=Asin(Bx-C)+D
I've been waiting for this video
Now let's talk about..
graphing..
..TRIGONOMETRIC FUNCTIONS.
EOC next week wish me luck guys
Mistake at 21:48 it should be 5 pi/2 not 4. Great Video regardless.
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I'm absolutley speechless .
14:17