Exact value of Trigonometric equation.

[ifomoe]

Homework Statement

$$1/\sqrt{}2 cos 15 - 1/\sqrt{}2 sin 15$$

The Attempt at a Solution

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

 

[tiny-tim]

Welcome to PF!

Homework Statement

$$1/\sqrt{}2 cos 15 - 1/\sqrt{}2 sin 15$$

The Attempt at a Solution

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

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?

 

[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.

 

[tiny-tim]

ifomoe brain good!

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.

Woohoo! ​

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

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

 

[ifomoe]

Yay!

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

 

[HallsofIvy]

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

 

[tiny-tim]

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

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