# Maths help Trig GRAPHSWatch

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#1
I have no clue how to draw this graph. I know that sine 180 is zero so I'm confused on how to draw the graph
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#2
Paper D Edexcel maths as paper
Q4a
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#3
This Paper:
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#4
bump
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1 year ago
#5
First off what does a sin(x) graph look like
Now if i multiply the entire function by the number i am stretching it vertically.
If I am multiplying the inside (the bit with x) and i am squashing the wave together so that things are more frequent.

Its just a transformation question.
Learn the transformation rules.
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1 year ago
#6
(Original post by Rolls_Reus_0wner)
I have no clue how to draw this graph. I know that sine 180 is zero so I'm confused on how to draw the graph
The period of the sine function sin(t degrees) is 360 degrees and so using 0<=t<=10 you need to work out how many cycles of the sine wave to draw for sin(180t degrees). The 1/2 factor is just a matter of scaling so the maxima and minima are 1/2 rather than 1.
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#7
(Original post by Dalek1099)
The period of the sine function sin(t degrees) is 360 degrees and so using 0<=t<=10 you need to work out how many cycles of the sine wave to draw for sin(180t degrees). The 1/2 factor is just a matter of scaling so the maxima and minima are 1/2 rather than 1.
I know about the 1/2 bit. I have never seen sin(180t) before. btw is there a way to calculate co-ordinates of this graph on a calculator
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1 year ago
#8
(Original post by Rolls_Reus_0wner)
I have no clue how to draw this graph. I know that sine 180 is zero so I'm confused on how to draw the graph
The inequality is from 0 ≤ t ≤ 10
The graph is 1/2 sin(180t).
Use transformation of functions to know that this is stretched vertically by 1/2 (so the range is -1/2 ≤ f(t) ≤ 1/2)
And that f(180t) means it's squeezed horizontally by a scale factor of 180.

So try plotting points for t = 1-10 to give a general idea of the graph
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1 year ago
#9
(Original post by Rolls_Reus_0wner)
I know about the 1/2 bit. I have never seen sin(180t) before. btw is there a way to calculate co-ordinates of this graph on a calculator
0<=t<=10 so plug in t=0 and t=10 into 180t and then that gives you the range of degrees that the sine function needs to be drawn and then from that you can work out the number of cycles from the period of the sine function.
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#10
(Original post by Dalek1099)
0<=t<=10 so plug in t=0 and t=10 into 180t and then that gives you the range of degrees that the sine function needs to be drawn and then from that you can work out the number of cycles from the period of the sine function.
If I put this graph into Desmos, it looks way different than the one in the markscheme.
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#11
bump
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1 year ago
#12
(Original post by Rolls_Reus_0wner)
If I put this graph into Desmos, it looks way different than the one in the markscheme.
What are you graphing on Desmos? The function sin(180t) simply contains a different argument 180t than you are used to so all that does is change the range of degrees that you have to graph the sine function for. The graph of sin(180t) for 0<=t<=10 is the same as the graph of sin(t) for 0<=t<=1800.
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#13
(Original post by Dalek1099)
What are you graphing on Desmos? The function sin(180t) simply contains a different argument 180t than you are used to so all that does is change the range of degrees that you have to graph the sine function for. The graph of sin(180t) for 0<=t<=10 is the same as the graph of sin(t) for 0<=t<=1800.
This is EXACTLY What i put into Desmos:

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#14
Ad you can see, it looks proper different from the markscheme
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1 year ago
#15
(Original post by Rolls_Reus_0wner)
This is EXACTLY What i put into Desmos:

Desmos works in Radians so that won't work as the question refers to degrees. To convert the Desmos graph to degrees use 1/2sin(pit) rather than 1/2sin(180t).
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#16
(Original post by Dalek1099)
Desmos works in Radians so that won't work as the question refer to degrees. To convert the Desmos graph to degrees use 1/2sin(pit) rather than 1/2sin(180t).
You legend ! Damn I thought it worked in degrees. Cheers man
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