Find the acute angle for a sine angle?

[wduk]

Hello

I have a sin theta = - 1/2sqrt(2)

And my book says that therefore sine phi = 1/sqrt(2).

This isn't obvious to me. Unless i am dealing with simple fractions of pi i don't quite know how i would have known that the acute angle was 1/sqrt(2). Can some one explain how this is worked out properly so i can use it for any future angle. Unless this is some kind of special angle and so it is generally known without calculating ?

 

[ksdhart2]

Well, based solely on the information included in your post, there's literally no way to know how they arrived at the answer of \(\displaystyle sin(\phi)\) as we don't know how the angles \(\displaystyle \theta\) and \(\displaystyle \phi\) are related. However, I'm willing to go out on a limb and assume that included in the problem was a right triangle, with the two unknown angles labelled \(\displaystyle \theta\) and \(\displaystyle \phi\) respectively. Assuming that is the correct interpretation, we can go fairly easily from \(\displaystyle sin(\theta)\) to \(\displaystyle sin(\phi)\) by using known properties of inverse trig functions and angles of triangles. Start with the given:

\(\displaystyle sin(\theta)=-\dfrac{1}{2} \sqrt{2}\)

Take the inverse sine of both sides:

\(\displaystyle sin^{-1}(sin(\theta))=sin^{-1} \left(-\dfrac{1}{2} \sqrt{2} \right) \implies \theta = \text{???}\)

Then you know that the three angles in a triangle must add up to 180 degrees, so:

\(\displaystyle \phi = 180^{\circ}-(90^{\circ}+\theta)\)

Or, if you prefer radians:

\(\displaystyle \phi = \pi-(90+\theta)\)

 

[wduk]

Okay i got the answer in the end after that post, thank you

I think you meant to put pi/2 not 90 in your radians equation for phi right ?

 

[ksdhart2]

Okay i got the answer in the end after that post, thank you

I think you meant to put pi/2 not 90 in your radians equation for phi right ?

Ah, yes, you're correct. I got a bit mixed up in my transition between degrees and radians. Glad you understood what I meant though.