Discover the Solution to Sin(arctan(x/4)) with Expert Guidance

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To solve Sin(arctan(x/4)), a right triangle is constructed with sides x and 4, where the angle θ corresponds to arctan(x/4). The tangent of angle θ is defined as x/4, leading to the identification of sin(θ) using the Pythagorean theorem to find the hypotenuse. After calculations, the expression for sin(θ) is derived as x√(x^2 + 16)/(x^2 + 16). This approach effectively illustrates the relationship between the angle and the sine function in the context of trigonometric identities. The solution is confirmed as correct.
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Sin(arctan(x/4))= ?

Been over 2 years since I've done some math, a little help please?
 
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Set up a right triangle with sides x and 4, so that the tangent of one of the angles is x/4, i.e., tanθ=x/4. Then θ=tan^{-1}(x/4) . From the drawing, figure out the value of sinθ.
 
It might help you if you draw a triangle and split the expression above into component parts.

First, how would you triangle look if you were to show what arctan(x/4) meant?
 
Ok I:

Drew a triangle in quadrant 1 to represent x/4 and labeled the angle A for random sake
then used Pythagorean theorem to find the hyp
after solving for sign and rationalizing I came up with:

SinA= (x(√(x^2)+16)/((x^2)+16) ------ √ ending after the first 16
sound right?
 
Yes, ##\sin(A)=x\frac{\sqrt{x^2+16}}{x^2+16}##.

ehild
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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