Question

Sin (2 arccos ) 3÷5=срочно ​

Answer 1

Объяснение:

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Answer 2

To solve this expression, let's break it down step by step.

First, we need to find the value of arccos(3/5). The arccos function gives us the angle whose cosine is equal to the given value. In this case, arccos(3/5) means finding the angle whose cosine is 3/5.

We can use the Pythagorean identity to find the missing side of a right triangle. Since cosine is adjacent over hypotenuse, we can set the adjacent side as 3 and the hypotenuse as 5, and solve for the opposite side.

Using the Pythagorean theorem: (opposite)^2 + (adjacent)^2 = (hypotenuse)^2 (opposite)^2 + 3^2 = 5^2 (opposite)^2 + 9 = 25 (opposite)^2 = 16 opposite = 4

So, the opposite side is 4. Therefore, arccos(3/5) is equal to the angle whose opposite side is 4 and hypotenuse is 5.

Next, we need to find sin(2 * arccos(3/5)). Using the double-angle identity for sine, sin(2θ) = 2 * sin(θ) * cos(θ).

Since we know the value of arccos(3/5), we can find sin(arccos(3/5)) by using the Pythagorean identity: sin(arccos(θ)) = √(1 - θ^2).

sin(arccos(3/5)) = √(1 - (3/5)^2) = √(1 - 9/25) = √(25/25 - 9/25) = √(16/25) = 4/5

Now, we can calculate sin(2 * arccos(3/5)):

sin(2 * arccos(3/5)) = 2 * sin(arccos(3/5)) * cos(arccos(3/5)) = 2 * (4/5) * (3/5) = (8/5) * (3/5) = 24/25

Therefore, sin(2 * arccos(3/5)) is equal to 24/25.