Homework Help: Exact value of Trigonometric equation.

1. Aug 9, 2008

ifomoe

1. The problem statement, all variables and given/known data
$$1/\sqrt{}2 cos 15 - 1/\sqrt{}2 sin 15$$

3. The attempt at a solution
I got $$\sqrt{}3/2$$.
If anyone could confirm this, I would appreciate it. Thanks.

2. Aug 9, 2008

tiny-tim

Welcome to PF!

Hi ifomoe! Welcome to PF!

(have a square-root: √ )

(I assume you mean (1/√2)cos15 - (1/√2)sin15?)

Nooo … you've used + instead of -

Which formula did you use?

3. Aug 9, 2008

ifomoe

Thank you.
Yes. I meant it like that: (1/√2)cos15 - (1/√2)sin15.
I used the reduction formula but I just re-did it and got 1/2.
Please, don't tell me I'm wrong again. My brain can't handle much more of this.

4. Aug 9, 2008

tiny-tim

ifomoe brain good!

Woohoo!

(btw, what's the "reduction formula"?

I used sin(45 - 15) = sin45 cos15 - cos45 sin15. )

5. Aug 9, 2008

ifomoe

Yay!

I think my lecturer said that another name for reduction formula is auxiliary angle formula.
If I'm not mistaken.

6. Aug 9, 2008

HallsofIvy

$$sin(x/2)= \sqrt{\frac{1}{2}(1- cos(x))}$$
so
$$sin(15)= \sqrt{\frac{1}{2}(1- cos(30))}= \sqrt{\frac{1}{2}\left(1- \frac{\sqrt{3}}{2}\right)}$$

7. Aug 9, 2008

tiny-tim

Or (cosθ - sinθ)/√2 = √2 sin45 sin(45 - θ) = sin(45 - θ).

Last edited: Aug 9, 2008