Question

Let sin x = .93 and cos y = .29. Which is x + y?

Answer 1

Let $\sin x = 0.93$ and $\cos y = 0.29$. Which is $x + y$?

Solution.

Consider $\sin x = 0.93$. Take the inverse sine of both sides:

$$x = (-1)^n \sin^{-1} 0.93 + \pi n$$

Then

$$x_1 = \sin^{-1} 0.93 + 2\pi n, \quad n \in \mathbb{Z}$$ $$x_2 = \pi - \sin^{-1} 0.93 + 2\pi n, \quad n \in \mathbb{Z}$$

Consider $\cos y = 0.29$. Take the inverse cosine of both sides:

$$y = \pm \cos^{-1} 0.29 + 2\pi n, \quad n \in \mathbb{Z}$$

Then

$$y_1 = \cos^{-1} 0.29 + 2\pi n, \quad n \in \mathbb{Z}$$ $$y_2 = -\cos^{-1} 0.29 + 2\pi n, \quad n \in \mathbb{Z}$$

Consider $x$ and $y$ on the interval $[0, 2\pi]$. So

$$x + y = \sin^{-1} 0.93 \pm \cos^{-1} 0.29 \quad \text{or} \quad x + y = \pi - \sin^{-1} 0.93 \pm \cos^{-1} 0.29$$

Answer:

$$x + y = \sin^{-1} 0.93 \pm \cos^{-1} 0.29 \quad \text{or} \quad x + y = \pi - \sin^{-1} 0.93 \pm \cos^{-1} 0.29$$