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trig question

x =sint y =tant

express y in terms of x

so y= (sint)/cost

so y=x/cost

and if you take x =sint and differentiate bswrt t

you get dx/dt = cost

im not really sure where to go from here :/
Reply 1
Avatar for Notnek
Notnek
21
Original post by TPJY
x =sint y =tant

express y in terms of x

so y= (sint)/cost

so y=x/cost

and if you take x =sint and differentiate bswrt t

you get dx/dt = cost

im not really sure where to go from here :/

No need to differentiate.

Can you express cos(t) in terms of sin(t)?
Reply 2
Avatar for TenOfThem
TenOfThem
18
Original post by TPJY
x =sint y =tant

express y in terms of x

so y= (sint)/cost

so y=x/cost



you have not done the first part yet as you still have t in your answer

can you turn cost into some function of x?
Original post by TPJY
x =sint y =tant

express y in terms of x

so y= (sint)/cost

so y=x/cost

and if you take x =sint and differentiate bswrt t

you get dx/dt = cost

im not really sure where to go from here :/


I think you might need a more explicit prompt.

Can you make use of the fact that:

sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1
what exam board are you on
Reply 5
Avatar for TPJY
TPJY
OP
10
Original post by notnek
No need to differentiate.

Can you express cos(t) in terms of sin(t)?


i was gunna go along the lines of cos(t-90) = sin t but from what mr m suggested then i can get cos t =(1-x^2)^1/2?

Original post by Mr M
I think you might need a more explicit prompt.

Can you make use of the fact that:

sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1
Reply 6
Avatar for Notnek
Notnek
21
Original post by TPJY
i was gunna go along the lines of cos(t-90) = sin t but from what mr m suggested then i can get cos t =(1-x^2)^1/2?

Correct :smile:
Reply 7
Avatar for TPJY
TPJY
OP
10
safe thanks :smile:

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