Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

#1 2006-06-24 10:00:11

fusilli_jerry89
Full Member

Offline

Trig

does anyone know how to get this question? I got pi/2 and 3pi/3.

√2 sin -sin = 0    Solve for values of sin. 0<x<2pi

#2 2006-09-01 19:31:10

John E. Franklin
Star Member

Offline

Re: Trig

How about 0.707


igloo myrtilles fourmis

#3 2006-09-01 21:12:49

mathsyperson
Moderator

Offline

Re: Trig

Forget the sins for the moment and solve it as if it was a simple quadratic equation.

√2 sinx -sinx = 0
sinx(√2 sinx -1) = 0
sinx = 0 or (√2 sinx -1) = 0

If (√2 sinx -1) = 0, then √2 sinx = 1 and so sinx = 1/√2.

So now we have sinx = 0 or 1/√2.
Therefore, x = 0, π/4, 3π/4, π, and 2π.
0 and 2π may not be answers, depending on whether the limits of the range are included.


Why did the vector cross the road?
It wanted to be normal.

#4 2006-09-01 23:58:49

John E. Franklin
Star Member

Offline

Re: Trig

Right, because .707 is 1/√2.


igloo myrtilles fourmis

#5 2006-09-02 00:01:19

John E. Franklin
Star Member

Offline

Re: Trig

Also half of the square root of 2 is the same as 1 over the square root of 2.


igloo myrtilles fourmis

Board footer

Powered by FluxBB