Please explain the steps in finding the exact value of sine 75 degrees and cosine 75 degrees.

Question

mathstudent55 · Answer

Have you learned the sine and cosine of a sum?

anonymous · Answer

yes, but I need to understand how they result comes to square root of something over something.

anonymous · Answer

couldn't you use sine and cosine of a difference though? @mathstudent55

mathstudent55 · Answer

$sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $

mathstudent55 · Answer

How would you express 75 degrees as a difference?

anonymous · Answer

135-60?

anonymous · Answer

IS it only applicable in one quadrant? I thought you can do it from all 4 quadrants.

But it should be the same as the formula you posted but with a minus operant

anonymous · Answer

$$\sin(\alpha-\beta)=\sin \alpha*\cos \beta-\sin \beta*\cos \alpha)$$

anonymous · Answer

like that

anonymous · Answer

It's one of the trig additional formulas

anonymous · Answer

http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

mathstudent55 · Answer

I would say 75 deg = 90 deg - 15 deg, but 15 deg is not one of the basic angles for which we know the sine and cosine.
That's why I'd use a sum: 75 deg = 45 deg + 30 deg.
45 deg and 30 deg are famous angles that come from the 45-45-90 and 30-60-90 triangles. We do know the exact values of the sine and cosine of 45 deg and 30 deg.

anonymous · Answer

ohh i see what your saying. Yeah you could do it that way.

You can take it from here

mathstudent55 · Answer

$sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $

$\sin 75^o$ 

$= \sin(45^o + 30^o)$

$= \sin 45^o \cos 30^o + \cos 45^o \sin 30^o $

Now substitute the exact values for the sines and cosines in the line above and simplify.

mathstudent55 · Answer

For the cosine, use this identity:

$ \cos(\alpha + \beta) = \cos \alpha \cos \beta − \sin \alpha \sin \beta $