Trigonometric Identities

Step 1: Start with the given identity: Step 2: Recall the identity for : Set , so , and the identity becomes: Step 3: Since , we see that both sides are equal, proving the identity. Thus, the equation holds.

Step 1: Understanding the Concept:
This question requires proving an identity involving inverse trigonometric functions.
The relationship between and can be established using a right-angled triangle and the fundamental trigonometric identity .
Step 2: Key Formula or Approach:
Let . This implies .
We will use the identity (for the principal value range of , which is , is non-negative).
From , we can find in terms of .
Step 3: Detailed Explanation:
Let's start by assuming .
By the definition of the inverse sine function, this means: Now, we use the Pythagorean identity .
We can rearrange this to solve for : Substituting into the equation: Taking the square root of both sides gives: (We take the positive root because the range of is , and in this interval, ).
Now, we can express using the inverse cosine function: Since we initially let , we can equate the two expressions for : Step 4: Final Answer:
We have successfully shown that both sides of the equation are equal to the same angle , thus proving the identity.

Step 1: Understanding the Concept:
The principal value of an inverse trigonometric function is the value that lies within its defined principal value branch.
For , the principal value branch is .
Step 2: Detailed Explanation:
Let .
By the definition of an inverse function, this implies:
We know that .
So, .
We know that .
Since , we can write:
So, .
Step 3: Final Answer:
We must check if this value lies in the principal value range of , which is .
Since and , the value is valid.
Therefore, the principal value of is .

Step 1: Understanding the Concept:
The problem requires simplifying a complex trigonometric expression involving an inverse function. The key is to simplify the argument of the function using trigonometric identities. The identities for are particularly useful.
Step 2: Key Formula or Approach:
1. Use the identities: 2. Substitute these into the expression and simplify the square roots, paying attention to the signs based on the given interval for .
3. Simplify the resulting fraction inside the function.
4. Use the property for in the appropriate range.
Step 3: Detailed Explanation:
Let's simplify the terms under the square roots. The problem states that . This means .
In this interval , both and are positive. Also, .
Therefore, is positive, and is also positive.
So we can remove the absolute value signs: Now, substitute these into the main expression's argument: Simplify the numerator and the denominator: So the original expression becomes: Since , which is within the principal value range of (which is ), we can simplify this to: Step 4: Final Answer:
We have proven that for .

Step 1: Understanding the Concept:
The principal value of an inverse trigonometric function, , is the unique angle in the range such that . For negative inputs, we use the identity .
Step 2: Key Formula or Approach:
1. Use the identity: .
2. Find the value of .
3. Substitute this value back into the identity to find the principal value.
Step 3: Detailed Explanation or Calculation:
We need to find the principal value of .
Using the identity, we have: We know that the angle in for which is .
So, .
Substituting this back: The value lies in the principal value range of , which is .
Step 4: Final Answer:
The principal value of is .