I take notes and study but nowhere in my lesson do i see anything about this?
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You also need this equation, which is always true:
(sin t)^2 + (cos t)^2 = 1
So now you have two equations, and two variables (sin t and cos t) so you can solve:
3 sin t − cos t = 0
=> 3 sin t = cos t
So we can substitute this into the other equation:
(sin t)^2 + (3 sin t)^2 = 1
etc.
You should get two possible values of sin t (one positive, one negative), and a corresponding cos t value for each.
(sin t)^2 + (cos t)^2 = 1
So now you have two equations, and two variables (sin t and cos t) so you can solve:
3 sin t − cos t = 0
=> 3 sin t = cos t
So we can substitute this into the other equation:
(sin t)^2 + (3 sin t)^2 = 1
etc.
You should get two possible values of sin t (one positive, one negative), and a corresponding cos t value for each.
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UGH...tan t = 1/3 ---> t = arctan (1/3) + nπ or if tan t = 1 / 3 then sin t = ± 1 / √(1+3253)
and cos t = ± 3 / √10
and cos t = ± 3 / √10
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