any ideas? sin[arccos(-sqrt[3]/2) + arctan(-sqrt[3]/3)] (easy angles)

[hellawowser]

. . . . .\(\displaystyle \sin\left[\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\, +\, \arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right]\)

so I know these are easy angles and we all know the values, but what if they weren't and I couldn't use a calculator? My teacher said it can be solved using sin(a+b) = sina cosb + sinb cosa, but I'm stuck.

. . . . .\(\displaystyle \sin\left(a\, +\, b\right)\, =\, \sin\left(\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\, +\, \arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right)\)

. . . . . . .\(\displaystyle =\, \sin\left(\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\right)\, \cos\left(\arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right)\, +\, \sin\left(\arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right)\, \cos\left(\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\right)\)

Then using the inverse functions properties and fundamental formulas I could only calculate sine and cosine of a , that is +/-1/4 and -sqrt3/2 respectively. But i have no idea what to do with sin(arctan) and cos(arctan)

 

[lev888]

If arctan is an "easy angle", it should be easy to calculate sin(easy angle). Not sure why the second expression is more difficult than the first.

 

[Dr.Peterson]

. . . . .\(\displaystyle \sin\left[\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\, +\, \arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right]\)

. . . . .\(\displaystyle \sin\left(a\, +\, b\right)\, =\, \sin\left(\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\, +\, \arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right)\)

. . . . . . .\(\displaystyle =\, \sin\left(\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\right)\, \cos\left(\arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right)\, +\, \sin\left(\arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right)\, \cos\left(\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\right)\)

Then using the inverse functions properties and fundamental formulas i could only calculate sin and cos of a , that is +-1/4 and -sqrt3/2 respectively. But i have no idea what to do with sin(arctan) and cos(arctan)
so i know these are easy angles and we all know the values,but what if they weren't and i couldn't use a calculator? My teacher said it can be solved using sin(a+b) = sina cosb + sinb cosa, but i'm stuck.

As I understand it, you are not asking how to do the problem as it stands without a calculator, but if the angles were not nice ones. For example, what if you wanted to calculate an expression for sin[arccos(x) + arctan(y)], or sin[arccos(1/3) + arctan(2)]?

Have you never seen a problem asking you to simplify sin(arctan(y)), or to evaluate, say, sin(arctan(2))? This is the sine of an angle whose tangent is y (or 2). You can draw a triangle with that ratio and solve for the sine; or you can use identities to relate the sine to the tangent (perhaps by way of the cosecant, for example).

Give it a try, and show us what you can do.