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How do you evaluate #sin(tan^-1(12/5))#?

Answer 1

#sin (tan^-1(12/5)) =12/13#

#sin (tan^-1(12/5))#. Let #tan^-1(12/5) = theta :. tan theta = 12/5# Since #tan^-1# exists in 1st quadrant & 4th quadrant and positive in 1st quadrant, #theta# is in 1st quadrant . we know #tan theta =p/b= 12/5 :. p=12 ; b=5 ; h= sqrt(p^2+b^2)=sqrt(12^2+5^2)=13 ; sin theta = p/h=12/13#. [p=perpendicular ; b = base; h= hypotenuse]

#:.sin (tan^-1(12/5)) =sin theta = 12/13# [Ans]

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Answer 2

Violet Anderson

To evaluate ( \sin(\tan^{-1}(12/5)) ), we first find the angle whose tangent is ( \frac{12}{5} ), which is approximately 67.38 degrees (in radians, approximately 1.175 radians). Then, we find the sine of this angle. So, ( \sin(\tan^{-1}(12/5)) ) is approximately 0.9726.

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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.