Write the trigonometric expression as an algebraic expression.

1.) SIN(ARCSIN X+ARCCOS X)
ANSWER: 1
2.) SIN(ARCTAN 2X-ARCCOS X)
ANSWER: 2x²-SquareRoot of 1-x²/Square
Root of 4x²+1.

To express the given trigonometric expressions as algebraic expressions, we can use the identities and properties of trigonometric functions.

1.) SIN(ARCSIN X+ARCCOS X)

We know that SIN(ARCSIN X) = X and SIN(ARCCOS X) = SquareRoot of (1 - X²)

Using these identities, the given expression can be simplified as follows:

SIN(ARCSIN X+ARCCOS X) = SIN(ARCSIN X) * COS(ARCCOS X) + COS(ARCSIN X) * SIN(ARCCOS X)

Since COS(ARCCOS X) = X and SIN(ARCSIN X) = X, the expression becomes:

X * X + COS(ARCSIN X) * SIN(ARCCOS X)

Now, we know that COS(ARCSIN X) = SquareRoot of (1 - X²) and SIN(ARCCOS X) = SquareRoot of (1 - X²)

Substituting these values, the expression becomes:

X * X + SquareRoot of (1 - X²) * SquareRoot of (1 - X²)

Using the identity a * a = a², we can simplify further:

X * X + SquareRoot of ((1 - X²) * (1 - X²))

Now, we can simplify the expression (1 - X²) * (1 - X²) using the algebraic formula for square of a binomial.

(1 - X²) * (1 - X²) = 1 - 2X² + X⁴

Substituting this value back into the expression, we get:

X * X + SquareRoot of (1 - 2X² + X⁴)

Since 1 - 2X² + X⁴ is a perfect square trinomial, we can express it as the square of a binomial:

X * X + SquareRoot of ((X² - 1)²)

Now, the expression becomes:

X * X + X² - 1

Combining like terms, we get:

2X² - 1

Therefore, the algebraic expression for SIN(ARCSIN X+ARCCOS X) is 2X² - 1.

2.) SIN(ARCTAN 2X-ARCCOS X)

We'll follow a similar process to express this trigonometric expression as an algebraic expression.

Using the identity SIN(ARCTAN X) = X / SquareRoot of (1 + X²), the expression can be expanded as:

SIN(ARCTAN 2X-ARCCOS X) = SIN(ARCTAN 2X) * COS(ARCCOS X) - COS(ARCTAN 2X) * SIN(ARCCOS X)

Since SIN(ARCTAN 2X) = 2X / SquareRoot of (1 + (2X)²), and COS(ARCTAN 2X) = 1 / SquareRoot of (1 + (2X)²), and using the previously mentioned identities for SIN(ARCCOS X) and COS(ARCCOS X), we can substitute these values into the expression.

After substituting and simplifying, the expression becomes:

(2X / SquareRoot of (1 + (2X)²)) * X - (1 / SquareRoot of (1 + (2X)²)) * SquareRoot of (1 - X²)

Combining like terms and further simplifying, we get:

2X² - SquareRoot of (1 - X²) / SquareRoot of (4X² + 1)

This can be written as:

2X² - SquareRoot of (1 - X²) / SquareRoot of (4X² + 1)

Therefore, the algebraic expression for SIN(ARCTAN 2X-ARCCOS X) is 2X² - SquareRoot of (1 - X²) / SquareRoot of (4X² + 1).

1.) To simplify the expression SIN(ARCSIN X+ARCCOS X), we can use the identities:

- SIN(ARCSIN θ) = θ
- COS(ARCCOS θ) = θ

So, applying these identities, we get:

SIN(ARCSIN X+ARCCOS X) = X + X = 2X

Therefore, the algebraic expression is 2X.

2.) To simplify the expression SIN(ARCTAN 2X-ARCCOS X), we can use the identities:

- SIN(ARCTAN θ) = θ/√(1+θ²)
- COS(ARCCOS θ) = θ

So, applying these identities, we get:

SIN(ARCTAN 2X-ARCCOS X) = (2X)/√(1+(2X)²) - X

To simplify further, we can rationalize the denominator:

= (2X)/√(1+4X²) - X * (√(1+4X²))/(√(1+4X²))

Now, we can combine the terms:

= (2X- X√(1+4X²))/(√(1+4X²))

Therefore, the algebraic expression is (2X- X√(1+4X²))/(√(1+4X²)).

1.) Apply the following method you used in #2! You should notably understand this already! Goodness grief!

2.) let A = arctan x
let B = arccos x

then we are looking for
sin(A + B) .

now,
tan A = x ,
sin A = x/√(x^2+1)
cos A = 1/√(x^2+1) ... create a right triangle where the opposite side is x, and the adjacent side is 1 ... the hypotenuse is √(x^2+1)

cos B = x
sin B = √(1 - x^2)

meanwhile
sin(A + B)
= sin A cos B + cos A sin B
= x/√(x^2+1) * x + 1/√(x^2+1) * √(1 - x^2)
= [x^2 + √(1 - x^2)] / (x^2 + 1) .