calculus and precalculus

Question

simplify answer, no calculator and use radicals if needed and rationalize denominator:  

sin(sin^-1 * 1/2 + tan^-1 * -5)

Answer 1

Let θ = arcsin(1/2) and φ = arctan(-5). Then sinθ = 1/2 and tanφ = -5. It is implied that A is in Quadrant I and B is in Quadrant IV by the range of the arcsin and arctan respectively.

We can use the fact that sin(θ + φ) = sinθ * cosφ + cosθ * sinφ.

If sinθ = 1/2 in QI, then cosθ = √3/2, because θ=30° .

If tanφ = -5 in QIV, then we can draw a triangle with the side opposite of φ equal to -5 and the adjacent side equal to 1, so that tanφ = -5/1 = -5. Then the hypotenuse would be √(5²+1²) = √26. From that same triangle, we see that sinφ = -5/√26 and cosφ = 1/√26 .

Thus, sin(θ + φ) = sinθ * cosφ + cosθ * sinφ

= (1/2) * (1/√26) + (√3/2) * (-5/√26)

= 1 /( 2√26) - 5√3 / (2√26) = (1 - 5√3) / (2√26)

= √26 * (1 - √5√3) / (2 * 26) [rationalize the denominator]

= (√26 - 5√78) / 52