Inverse of trig functions
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Inverse of trig functions

[From: ] [author: ] [Date: 11-11-21] [Hit: ]
and part B: write sin(theta) as a function of x.im so confused any help would be great-In the triangle just read that tan θ = (x+3)/4.c² = 4² + (x+1)² = 16 + x² + 2x + 1 = x² + 2x + 17 (cant be factored).Thus c = √(x² + 2x + 17).sin θ = (x+ 4)/c = (x+4)/√(x² + 2x + 17) OR (x+4)/√(4² + (x+1)²)-From the triangle,So,......
Part A: use an inverse function to write theta as a function of x. and there's a triangle drawn out where 4 and (x+3) are the legs of the triangle and theta is the angle opposite of (x+3).

and part B: write sin(theta) as a function of x.
i'm so confused any help would be great

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In the triangle just read that tan θ = (x+3)/4. Hence θ = tan⁻¹ (x+3)/4

In the triangle use the Pythagorean Theorem to compute the hypotenuse c:

c² = 4² + (x+1)² = 16 + x² + 2x + 1 = x² + 2x + 17 (can't be factored).
Thus c = √(x² + 2x + 17).

sin θ = (x+ 4)/c = (x+4)/√(x² + 2x + 17) OR (x+4)/√(4² + (x+1)²)

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From the triangle, we see that:

tan(theta) = (x+3)/4

so: theta = arctan((x+3)/4)


The hypotenuse is equal to Sqrt(16 + (x+3)^2)

So, sin(theta) = (x+3)/Sqrt(16 + (x+3)^2)
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