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How do you simplify the expression #sin(2 arctan x) #?

Answer 1

Nolan Alexander

#sin(2 arctan x) ={2x}/{1+x^2}#

Let's assume by #arctan x# we mean the multivalued inverse, all the angles whose tangent is #x#.

The double angle formula for sine is

# sin (2 theta) = 2 sin theta cos theta #

#sin(2 arctan x) = 2 sin(arctan x) cos(arctan x)#

We can think of #arctan (x/1) # as a right triangle whose opposite is #x# and adjacent is #1#, so the hypotenuse is #sqrt{1+x^2}.#

The sign of sine and of cosine are each ambiguous when we know the tangent. But the sign of the product sine and cosine and the sign of the tangent (which is the quotient of sine and cosine) are the same.

# sin arctan x = text{opp}/text{hyp} = pm x/sqrt{1+x^2}#

# cos arctan x = text{opp}/text{hyp} = pm 1/sqrt{1+x^2}#

#sin(2 arctan x) = 2 sin(arctan x) cos(arctan x) = {2x}/{1+x^2}#

Despite the ambiguity of the sign of the factors, the sign of the sine of the double angle is not ambiguous.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.