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Math Help - How do I simplify inverse functions trig?

  1. #1
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    How do I simplify inverse functions trig?

    Okay, i don't really know where to start
    "simplify completely. your answers should not contain trigonometric or inverse trigonometric functions
    sin(Arcsin2/3+Arccos\sqrt3/2)
    So I think I should solve for Arcsin2/3 and Arcos\sqrt3/2, then add them to get one number and take the sin of that number.
    Would that be the steps?
    however, I don't know how to solve for Arcsin2/3 and Arccos\sqrt3/2
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  2. #2
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    Hello dorkymichelle
    Quote Originally Posted by dorkymichelle View Post
    Okay, i don't really know where to start
    "simplify completely. your answers should not contain trigonometric or inverse trigonometric functions
    sin(Arcsin2/3+Arccos\sqrt3/2)
    So I think I should solve for Arcsin2/3 and Arcos\sqrt3/2, then add them to get one number and take the sin of that number.
    Would that be the steps?
    however, I don't know how to solve for Arcsin2/3 and Arccos\sqrt3/2
    Get rid of inverse trig functions as quickly as you can - they're not nice! Use new letters to stand for them. So, let:
    x =\arcsin \tfrac23 \Rightarrow \sin x = \tfrac 23
    and
    y = \arccos\tfrac{\sqrt3}{2} \Rightarrow \cos y = \tfrac{\sqrt3}{2}
    That feels better already!

    Now, we want \sin (x+y), so that's \sin x \cos y + \cos x \sin y.

    We know \sin x and \cos y. Can you work out \cos x and \sin y, using Pythagoras? When you've done that, plug all the values in, and you're there.

    I make the answer \frac{2\sqrt3 + \sqrt5}{6}. Do you agree?

    Grandad
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