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Thread: 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
    $\displaystyle sin(Arcsin2/3+Arccos\sqrt3/2)$
    So I think I should solve for $\displaystyle Arcsin2/3$ and $\displaystyle Arcos\sqrt3/2$, 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 $\displaystyle Arcsin2/3$ and $\displaystyle Arccos\sqrt3/2$
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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
    $\displaystyle sin(Arcsin2/3+Arccos\sqrt3/2)$
    So I think I should solve for $\displaystyle Arcsin2/3$ and $\displaystyle Arcos\sqrt3/2$, 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 $\displaystyle Arcsin2/3$ and $\displaystyle Arccos\sqrt3/2$
    Get rid of inverse trig functions as quickly as you can - they're not nice! Use new letters to stand for them. So, let:
    $\displaystyle x =\arcsin \tfrac23 \Rightarrow \sin x = \tfrac 23$
    and
    $\displaystyle y = \arccos\tfrac{\sqrt3}{2} \Rightarrow \cos y = \tfrac{\sqrt3}{2}$
    That feels better already!

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

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

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

    Grandad
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