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) .