Anyone know how to do this trig problem?

[Schadenfroh]

your links are broke

 

[Shawn]

wtf? you guys killed my school's server.

 

[FrustratedUser]

Originally posted by: Shawn
wtf? you guys killed my school's server.

0wn3d!!

 

[TuxDave]

lol... this is one of those 1 second problems.

 

[Shawn]

Originally posted by: TuxDave
lol... this is one of those 1 second problems.

It says the answer is sin x, but wouldn't it be -sin x since it's an odd function?

 

[Kyteland]

Originally posted by: Shawn

Originally posted by: TuxDave
lol... this is one of those 1 second problems.

It says the answer is sin x, but wouldn't it be -sin x since it's an odd function?

No, sin repeats with a period of 2*pi.
sin(x) = sin(x+2*pi)

This means sin(3*pi-x) = sin(pi-x).

Sin has an inverse with a period of pi.
sin(x) = -sin(x+pi)

This means sin(pi-x) = -sin(-x) = sin(x)

As a secondary proof you can use the addition formula for sin:
sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y)

sin(3*pi-x) = sin(3*pi)*cos(-x)+cos(3*pi)*sin(-x)
= 0*cos(-x)+(-1)*sin(-x)
= -sin(-x)
= sin(x)

 

[StevenYoo]

Originally posted by: Kyteland

Originally posted by: Shawn

Originally posted by: TuxDave
lol... this is one of those 1 second problems.

It says the answer is sin x, but wouldn't it be -sin x since it's an odd function?

No, sin repeats with a period of 2*pi.
sin(x) = sin(x+2*pi)

This means sin(3*pi-x) = sin(pi-x).

Sin has an inverse with a period of pi.
sin(x) = -sin(x+pi)

This means sin(pi-x) = -sin(-x) = sin(x)

As a secondary proof you can use the addition formula for sin:
sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y)

sin(3*pi-x) = sin(3*pi)*cos(-x)+cos(3*pi)*sin(-x)
= 0*cos(-x)+(-1)*sin(-x)
= -sin(-x)
= sin(x)

that's what i was gonna say