- #1

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so f(t)=8sint+tant+c

4=8sin(pi/3)+tan(pi/3)+c

c=-3^1/2

so f(t)=8sint+tant-3^1/2 is that right

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- Thread starter Rasine
- Start date

- #1

- 205

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so f(t)=8sint+tant+c

4=8sin(pi/3)+tan(pi/3)+c

c=-3^1/2

so f(t)=8sint+tant-3^1/2 is that right

- #2

- 2,063

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You need to re-evaluate the values of sin and tan at pi/3.

- #3

- 205

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thank you very much

- #4

- 205

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4sqroot3+sqroot3=5sqroot3

is that right?

- #5

- 2,063

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But for the sake of clarity, use something like sqroot(3)/2 instead of sqroot3/2 when referring to [tex]\frac{\sqrt{3}}{2}[/tex].

- #6

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ok. so now i have (5sqroot(3)/2)+c=4

then c=4-5sqroot(3)/2 so that would be -sqroot3

is that right

then c=4-5sqroot(3)/2 so that would be -sqroot3

is that right

- #7

- 2,063

- 2

Why the 2 in the denominator?then c=4-5sqroot(3)/2

so that would be -sqroot3

I don't understand. You've already solved for c, so...

f(t) = 8sin(t) + tan(t) + 4-5sqroot(3)

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