Question

(arcsin x)(arccos x)= -π2/9π2- пи в квадрате

Answer 1

Решение на фотографии

Answer 2

Let's simplify the expression (arcsin x)(arccos x) and see if it matches the given result.

We know that for any real number x, the following trigonometric identity holds:

arcsin(x) + arccos(x) = π/2

Now, let's use this identity to rewrite the expression (arcsin x)(arccos x):

(arcsin x)(arccos x) = (arcsin x) * (π/2 - arcsin x)

To simplify further, let's use the property of trigonometric functions that states:

arcsin(x) + arcsin(-x) = π

So, rearranging the terms in the expression:

(arcsin x)(arccos x) = (arcsin x) * (π/2 - arcsin x) = (arcsin x) * (arcsin(-x) + π/2)

Now, we have a product of two arcsin functions. We can simplify it using another trigonometric identity:

sin(arcsin(y)) = y

So, the expression becomes:

(arcsin x)(arccos x) = (arcsin x) * (arcsin(-x) + π/2) = (arcsin x) * (x + π/2)

Finally, to find the value of the expression, we need to remember that arcsin and arccos functions give values in the range [-π/2, π/2] and [0, π], respectively.

Now, let's evaluate the expression:

(arcsin x)(arccos x) = (arcsin x) * (x + π/2)

Since arcsin(x) gives a value in the range [-π/2, π/2], and x is a real number, the only possibility for the product of these two values to be equal to the given result (-π^2 / 9 * π^2) is when x is equal to zero.

Therefore, the expression (arcsin x)(arccos x) equals zero when evaluated at x = 0, and it does not match the given result of (-π^2 / 9 * π^2).