How do you simplify the expression $cos [arcsin (x)] \cdot sin [arccos (x)]$ $cos[arcsin(x)] * sin[arccos(x)]$ ?

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How do you simplify the expression $cos [arcsin (x)] \cdot sin [arccos (x)]$ $cos[arcsin(x)] * sin[arccos(x)]$ ?

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Answer 1 · 2016-10-10T00:24:15

$cos (arcsin (x)) \cdot sin (arccos (x)) = 1 - x^{2}$ $cos(arcsin(x))*sin(arccos(x))=1-x^2$

Explanation:

Let's draw a right triangle.

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In this triangle, we've chosen a leg of length $x$ $x$ and a hypotenuse of length $1$ $1$ , meaning $sin (a) = cos (b) = \frac{x}{1} = x$ $sin(a) = cos(b) = x/1 = x$ . Because of that, we know that $a = arcsin (x)$ $a = arcsin(x)$ and $b = arccos (x)$ $b = arccos(x)$ . Additionally, we can use the Pythagorean theorem to find the length of the remaining leg as $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$

We can now use this triangle to find the product.

$cos (arcsin (x)) \cdot sin (arccos (x)) = cos (a) \cdot sin (b)$ $cos(arcsin(x))*sin(arccos(x)) = cos(a)*sin(b)$

$= \frac{\sqrt{1 - x^{2}}}{1} \cdot \frac{\sqrt{1 - x^{2}}}{1}$ $=sqrt(1-x^2)/1 * sqrt(1-x^2)/1$

$= 1 - x^{2}$ $=1-x^2$

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How do you simplify the expression $cos [arcsin (x)] \cdot sin [arccos (x)]$ $cos[arcsin(x)] * sin[arccos(x)]$ ?

1 Answer

Explanation:

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