cos(arcsin(1 + x))
Let θ = arcsin(1 + x), and a, b, c be the adjacent, opposite and hypotenuse sides respectively.
In our right triangle, b = 1 + x is now the side opposite of the angle θ, with c = 1 being the hypotenuse. This satisfies the relationship sin(θ) = b / c = (1 + x) / 1, as the sine is the opposite side divided by the hypotenuse.
Trough Pythagoras we can obtain our adjacent side:
a² + b² = c² ⇒ a = √(c² − b²) = √(1 − (1 + x)²).
Now cos(θ) = a / c = √(1 − (1 + x)²) / 1, as the cosine is the adjacent side divided by the hypotenuse.
And thus:
cos(arcsin(1 + x)) = √(1 − (1 + x)²) / 1 = √(− x² − 2x)
Let θ = arcsin(1 + x), and a, b, c be the adjacent, opposite and hypotenuse sides respectively.
In our right triangle, b = 1 + x is now the side opposite of the angle θ, with c = 1 being the hypotenuse. This satisfies the relationship sin(θ) = b / c = (1 + x) / 1, as the sine is the opposite side divided by the hypotenuse.
Trough Pythagoras we can obtain our adjacent side:
a² + b² = c² ⇒ a = √(c² − b²) = √(1 − (1 + x)²).
Now cos(θ) = a / c = √(1 − (1 + x)²) / 1, as the cosine is the adjacent side divided by the hypotenuse.
And thus:
cos(arcsin(1 + x)) = √(1 − (1 + x)²) / 1 = √(− x² − 2x)
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= sqrt(- x^2 - 2x)