# Sketch the graph - Polar curves

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Sketch the graph of r = sec(theta - pi/4)

Used a table of values from theta = 0 to 2 pi but didn't get the straight line graph required.

Please advise how to sketch the graph of this type.

Used a table of values from theta = 0 to 2 pi but didn't get the straight line graph required.

Please advise how to sketch the graph of this type.

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Sketch the graph of r = sec(theta - pi/4)

Used a table of values from theta = 0 to 2 pi but didn't get the straight line graph required.

Please advise how to sketch the graph of this type.

**golgiapparatus31**)Sketch the graph of r = sec(theta - pi/4)

Used a table of values from theta = 0 to 2 pi but didn't get the straight line graph required.

Please advise how to sketch the graph of this type.

What did you get for the table?

Last edited by mqb2766; 1 month ago

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(Original post by

https://www.desmos.com/calculator/r1umtv4egb

What did you get for the table?

**mqb2766**)https://www.desmos.com/calculator/r1umtv4egb

What did you get for the table?

r = 1.41, 1, 1.41, infinity, -1.41, -1, -1.41, infinity, 1.41

I found it difficult to plot these points (r, theta). Couldn't handle the negatives and infinity very well.

But I found another way to get the answer:

I wrote rcos(theta - pi/4 ) = 1

then used the cos(A-B) formula and the relations x = r cos theta and y = r sin theta to get x+y=sqrt 2 which is the line

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θ = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi

r = 1.41, 1, 1.41, infinity, -1.41, -1, -1.41, infinity, 1.41

I found it difficult to plot these points (r, theta). Couldn't handle the negatives and infinity very well.

But I found another way to get the answer:

I wrote rcos(theta - pi/4 ) = 1

then used the cos(A-B) formula and the relations x = r cos theta and y = r sin theta to get x+y=sqrt 2 which is the line

**golgiapparatus31**)θ = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi

r = 1.41, 1, 1.41, infinity, -1.41, -1, -1.41, infinity, 1.41

I found it difficult to plot these points (r, theta). Couldn't handle the negatives and infinity very well.

But I found another way to get the answer:

I wrote rcos(theta - pi/4 ) = 1

then used the cos(A-B) formula and the relations x = r cos theta and y = r sin theta to get x+y=sqrt 2 which is the line

sec = hyp/adj

So for an adjacent side of 1, the hypotenuse is r hence it's a straight line. When theta=pi/4, r=1.

Last edited by mqb2766; 1 month ago

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(Original post by

Yes the simple trig is

sec = hyp/adj

So for an adjacent side of 1, the hypotenuse is r hence it's a straight line. When theta=pi/4, r=1.

**mqb2766**)Yes the simple trig is

sec = hyp/adj

So for an adjacent side of 1, the hypotenuse is r hence it's a straight line. When theta=pi/4, r=1.

I don't understand the part "for an adjacent side of 1, the hypotenuse is r hence it's a straight line". How do you know? I had to do the algebraic manipulation to get it in the form x +y = k to know it's a straight line

Also, do you have general advice on how to plot polar curves?

Thanks a lot for your help

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(Original post by

Thank you.

I don't understand the part "for an adjacent side of 1, the hypotenuse is r hence it's a straight line". How do you know? I had to do the algebraic manipulation to get it in the form x +y = k to know it's a straight line

Also, do you have general advice on how to plot polar curves?

Thanks a lot for your help

**golgiapparatus31**)Thank you.

I don't understand the part "for an adjacent side of 1, the hypotenuse is r hence it's a straight line". How do you know? I had to do the algebraic manipulation to get it in the form x +y = k to know it's a straight line

Also, do you have general advice on how to plot polar curves?

Thanks a lot for your help

https://www.desmos.com/calculator/m1phxncbv7

Note if it was just

r = sec(theta)

This would be a vertical line passing through the x-axis at x=1. This is because it corresponds to a right angled triangle where r is the hypotenuse, x is the adjacent base and theta the angle. The base 1 is fixed and as theta varies, the hypotenuse traces out a vertical line. For simple trig functions, it's worth thinking about the geometry as well as algebra.

Yours is similar, except the "base" is rotated by pi/4.

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(Original post by

I've updated the diagram a bit

https://www.desmos.com/calculator/m1phxncbv7

Note if it was just

r = sec(theta)

This would be a vertical line passing through the x-axis at x=1. This is because it corresponds to a right angled triangle where r is the hypotenuse, x is the adjacent base and theta the angle. The base 1 is fixed and as theta varies, the hypotenuse traces out a vertical line. For simple trig functions, it's worth thinking about the geometry as well as algebra.

Yours is similar, except the "base" is rotated by pi/4.

**mqb2766**)I've updated the diagram a bit

https://www.desmos.com/calculator/m1phxncbv7

Note if it was just

r = sec(theta)

This would be a vertical line passing through the x-axis at x=1. This is because it corresponds to a right angled triangle where r is the hypotenuse, x is the adjacent base and theta the angle. The base 1 is fixed and as theta varies, the hypotenuse traces out a vertical line. For simple trig functions, it's worth thinking about the geometry as well as algebra.

Yours is similar, except the "base" is rotated by pi/4.

The diagram also made it clear to me that my points that I was plotting were just not accurate enough.

Thank you

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