How can I solve these set of parametric equations in terms of t using just trig identities?

x = tan t, y = sin t.

Thanks so much for your help!

x = tan t, y = sin t.

Thanks so much for your help!

Original post by Silverwolf146

How can I solve these set of parametric equations in terms of t using just trig identities?

x = tan t, y = sin t.

Thanks so much for your help!

x = tan t, y = sin t.

Thanks so much for your help!

You should know the basic trig identity/definition for tan ...

Uh... what do you mean by solve? We have 2 equations and 3 unknowns - there is definitely not a unique solution.

If you want to eliminate t, well, what trig identities do you know? (Like seriously, write them all down, perhaps with reference to your notes - surely some will work)

If you want to eliminate t, well, what trig identities do you know? (Like seriously, write them all down, perhaps with reference to your notes - surely some will work)

(edited 1 month ago)

Original post by Silverwolf146

How can I solve these set of parametric equations in terms of t using just trig identities?

x = tan t, y = sin t.

Thanks so much for your help!

x = tan t, y = sin t.

Thanks so much for your help!

I can Help you handle this

To solve x=tantx = \tan tx=tant and y=sinty = \sin ty=sint, use the identity:

sin2t+cos2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1

Since tant=sintcost\tan t = \frac{\sin t}{\cos t}tant=costsint, solve for cost\cos tcost in terms of xxx:

x=ycost ⟹ cost=yxx = \frac{y}{\cos t} \implies \cos t = \frac{y}{x}x=costy⟹cost=xy

Substitute into the identity:

y2+(yx)2=1y^2 + \left(\frac{y}{x}\right)^2 = 1y2+(xy)2=1

This gives a relation between xxx and yyy

sin2t+cos2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1

Since tant=sintcost\tan t = \frac{\sin t}{\cos t}tant=costsint, solve for cost\cos tcost in terms of xxx:

x=ycost ⟹ cost=yxx = \frac{y}{\cos t} \implies \cos t = \frac{y}{x}x=costy⟹cost=xy

Substitute into the identity:

y2+(yx)2=1y^2 + \left(\frac{y}{x}\right)^2 = 1y2+(xy)2=1

This gives a relation between xxx and yyy

Original post by thejohnwatson

To solve x=tantx = \tan tx=tant and y=sinty = \sin ty=sint, use the identity:

sin2t+cos2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1

Since tant=sintcost\tan t = \frac{\sin t}{\cos t}tant=costsint, solve for cost\cos tcost in terms of xxx:

x=ycost ⟹ cost=yxx = \frac{y}{\cos t} \implies \cos t = \frac{y}{x}x=costy⟹cost=xy

Substitute into the identity:

y2+(yx)2=1y^2 + \left(\frac{y}{x}\right)^2 = 1y2+(xy)2=1

This gives a relation between xxx and yyy

sin2t+cos2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1

Since tant=sintcost\tan t = \frac{\sin t}{\cos t}tant=costsint, solve for cost\cos tcost in terms of xxx:

x=ycost ⟹ cost=yxx = \frac{y}{\cos t} \implies \cos t = \frac{y}{x}x=costy⟹cost=xy

Substitute into the identity:

y2+(yx)2=1y^2 + \left(\frac{y}{x}\right)^2 = 1y2+(xy)2=1

This gives a relation between xxx and yyy

Your post hasn't come out in a readable format, but please note for future reference that it's against forum rules to post full solutions to problems - please stick to hints, suggestions etc as per forum guidelines. Thanks.

Original post by Silverwolf146

x = tan t, y = sin t.

Thanks so much for your help!

To solve these two parametric equations in terms of t using trigonometric identities, you need to use the following method:

•

x = tan t is already in terms of t, so you don't need to do anything with it.

•

For y = sin t, you can use the fact that sin^2 t + cos^2 t = 1 to rewrite y in terms of x.

Sandro

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