1. ## Trigonometric Functions.. again

Hi...again.

I know I'm a bother but can you guys help me again?

I'm having trouble with these ones. Thanks!

1. find the exact solutions algebraically in the interval [0, 2pi)
sin 2x sin x = cos x. (2 is not an exponent on this one)

2. verify identity algebraically
cos^4 x - sin^4 x = cos 2x.

3. find all solutions of the equation in the interval [0, 2pi)
sin(x + pi/6) - sin(x - pi/6) = 1/2

2. Originally Posted by cuteisa89

2. verify identity algebraically
cos^4 x - sin^4 x = cos 2x.
Difference of two squares,

(cos^2 x + sin^2 x)(cos^2 x - sin^2 x)

The first factor is a Pythagorean Identity,
Thus,

cos^2 x - sin^2 x = cos 2x

By double angle.

3. Originally Posted by cuteisa89
3. find all solutions of the equation in the interval [0, 2pi)
sin(x + pi/6) - sin(x - pi/6) = 1/2
Add them (the sines) together to get,

2*cos(2x)*sin(pi/3) = 1/2

Thus,

2*cos(2x)*(1/2) = (1/2)

Thus,

cos(2x) = 1/2

4. Originally Posted by cuteisa89
1. find the exact solutions algebraically in the interval [0, 2pi)
sin 2x sin x = cos x. (2 is not an exponent on this one)
sin(2x)*sin(x) = cos(x)

2*sin(x)*cos(x)*sin(x) = cos(x)

2*sin^2(x)*cos(x) - cos(x) = 0

cos(x)*[2*sin^2(x) - 1] = 0

Thus
cos(x) = 0 ==> x = (pi)/2, 3(pi)/2
or
2*sin^2(x) - 1 = 0

2*sin^2(x) = 1

sin^2(x) = 1/2

sin(x) = (+/-)1/sqrt{2}

x = (pi)/4, 3(pi)/4, 5(pi)/4, 7(pi)/4

Thus
x = (pi)/4, (pi)/2, 3(pi)/4, 5(pi)/4, 3(pi)/2, 7(pi)/4

-Dan