Question

Arcsin + arccos

Answer 1

Ответ:

$arcsin\dfrac{1}{2}+arccos\dfrac{\sqrt2}{2}=\dfrac{\pi}{6}+\dfrac{\pi}{4}=\dfrac{2\pi +3\pi }{12}=\dfrac{5\pi }{12}$

Answer 2

The sum of the inverse trigonometric functions arcsin (also denoted as sin^(-1)) and arccos (also denoted as cos^(-1)) is not a standard trigonometric identity. However, you can calculate it as follows:

Let's say you want to find the value of arcsin(x) + arccos(x), where x is a real number between -1 and 1, inclusive.

1. Start with the identity: sin^2(θ) + cos^2(θ) = 1, which is true for all real values of θ.

2. Now, we can write sin^2(θ) = 1 - cos^2(θ) and cos^2(θ) = 1 - sin^2(θ).

3. Take the square root of both sides of each equation, remembering that sin(θ) and cos(θ) are non-negative for real values of θ between 0 and π.

   √(sin^2(θ)) = √(1 - cos^2(θ)) √(cos^2(θ)) = √(1 - sin^2(θ))

4. Since sin(θ) and cos(θ) are non-negative in this context, you can write:

   sin(θ) = √(1 - cos^2(θ)) cos(θ) = √(1 - sin^2(θ))

5. Now, let's substitute arcsin(x) for θ and simplify:

   sin(arcsin(x)) = √(1 - cos^2(arcsin(x))) = √(1 - x^2)

   cos(arcsin(x)) = √(1 - sin^2(arcsin(x))) = √(1 - x^2)

6. Now, you can rewrite arcsin(x) + arccos(x) as follows:

   arcsin(x) + arccos(x) = arcsin(x) + arcsin(x)

7. Combine the two arcsin terms:

   arcsin(x) + arcsin(x) = 2 * arcsin(x)

So, the sum of arcsin(x) and arccos(x) is equal to 2 * arcsin(x):

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

This is the expression for the sum of arcsin and arccos when x is a real number between -1 and 1, inclusive.