Question

Find tan⁡(sin−1⁡(2/5)+cos−1⁡(5/10)) =

(Make sure your answer is an algebraic expression with square roots but without trigonometric or inverse trignometric functions.)

Express in terms of x: sin⁡(2tan−1⁡(x)) =

Answer 1

For the first problem:

tan(sin⁻¹(2/5) + cos⁻¹(5/10))

Let's start by using the identity: cos⁻¹(a) = sin⁻¹(√(1-a²))

cos⁻¹(5/10) = sin⁻¹(√(1-(5/10)²)) = sin⁻¹(√(1/4)) = sin⁻¹(1/2)

Now we can substitute this back into the original expression:

tan(sin⁻¹(2/5) + cos⁻¹(5/10)) = tan(sin⁻¹(2/5) + sin⁻¹(1/2))

Next, we can use the identity: tan(a + b) = (tan(a) + tan(b))/(1 - tan(a)tan(b))

tan(sin⁻¹(2/5) + sin⁻¹(1/2)) = (2/5 + 1/2)/(1 - (2/5)(1/2))

Simplifying the numerator and denominator:

= (4/10 + 5/10)/(10/10 - 4/10) = 9/6

Reducing the fraction:

= 3/2

Therefore, tan(sin⁻¹(2/5) + cos⁻¹(5/10)) = 3/2.

For the second problem:

sin(2tan⁻¹(x)) = 2tan⁻¹(x)

Let θ = tan⁻¹(x)

Then we have:

sin(2θ) = 2tan(θ)/(1+tan²(θ))

= 2x/(1+x²)

Therefore, sin(2tan⁻¹(x)) = 2x/(1+x²).

Answer 2

Let's solve each problem step by step:

Find tan⁡(sin⁻¹(2/5)+cos⁻¹(5/10)):

We know that sin⁻¹(2/5) is an angle whose sine is 2/5, and cos⁻¹(5/10) is an angle whose cosine is 5/10 (or 1/2). Let's call these angles θ and φ, respectively. Then:

sinθ = 2/5 and cosφ = 1/2

We can use the Pythagorean identity to find sinφ:

sin²φ + cos²φ = 1

sin²φ + (1/2)² = 1

sin²φ = 3/4

sinφ = √(3/4) = √3/2

Now, we can use the sum formula for tangent:

tan(θ + φ) = (tanθ + tanφ) / (1 - tanθ tanφ)

tan(θ + φ) = (2/5 + √3/5) / (1 - (2/5)(√3/2))

tan(θ + φ) = (2√3 + 3) / (5√3 - 6)

Therefore, tan⁡(sin⁻¹(2/5)+cos⁻¹(5/10)) = (2√3 + 3) / (5√3 - 6).

Express in terms of x: sin⁡(2tan⁻¹(x)):

Let's draw a right triangle with opposite side x, adjacent side 1, and hypotenuse r:

/|

/ |

/__|

1 x

Then, we know that tan⁻¹(x) is an angle whose tangent is x. Therefore:

tan(tan⁻¹(x)) = x

We can use the Pythagorean theorem to find r:

r² = 1² + x²

r = √(1 + x²)

Now, we can find sin(2θ) using the double-angle formula:

sin(2θ) = 2 sinθ cosθ

sinθ = x/r = x/√(1 + x²)

cosθ = 1/r = 1/√(1 + x²)

Therefore:

sin⁡(2tan⁻¹(x)) = 2 sin(tan⁻¹(x)) cos(tan⁻¹(x))

sin⁡(2tan⁻¹(x)) = 2 (x/√(1 + x²)) (1/√(1 + x²))

sin⁡(2tan⁻¹(x)) = 2x / (1 + x²)