Write the trigonometric expression as an algebraic expression.
sin(arcsinx+arccosx)
sin arcsinx cos arccosx + cos arcsinx sin arccosx
= x*x + √(1-x^2) √(1-x^2)
= x^2 + 1-x^2
= 1
Naturally, since arcsinx + arccosx = pi/2
To express the trigonometric expression sin(arcsinx + arccosx) as an algebraic expression, let's start by using the identities involving arcsine and arccosine.
1. The identity arcsin(sinθ) = θ applies when -π/2 ≤ θ ≤ π/2.
2. The identity arccos(cosθ) = θ applies when 0 ≤ θ ≤ π.
Since arcsinx and arccosx are within the appropriate ranges, we can use these identities to simplify the expression.
First, let's simplify arcsinx + arccosx:
arcsinx + arccosx = θ + Φ, where θ = arcsinx and Φ = arccosx
Since arcsin(sinθ) = θ and arccos(cosθ) = θ, we can rewrite the sum as:
= sin(θ) + cos(θ)
Now, let's express sin(θ) + cos(θ) as an algebraic expression:
To do this, we'll use the Pythagorean Identity: sin²θ + cos²θ = 1.
Rearranging the terms, we have:
sin²θ = 1 - cos²θ
Taking the square root of both sides, we get:
sinθ = ±√(1 - cos²θ)
Therefore, sin(θ) + cos(θ) can be expressed as:
±√(1 - cos²θ) + cos(θ)
Replacing θ with arcsinx, we have:
±√(1 - cos²(arcsinx)) + cos(arcsinx)
Simplifying further, we have:
±√(1 - x²) + √(1 - x²)
Combining the square root terms, we get:
2√(1 - x²)
Hence, the expression sin(arcsinx + arccosx) can be expressed algebraically as:
2√(1 - x²)