Question

(1) Evaluate exactly tan (Arccos 2/3)

(2) Evaluate exactly cos (Arcsin 1 + Arccos 1/2)

(3) Determine arcsin 1/√2
Careful: that is arcsin and not Arcsin, so there are going to
be infinitely many answers. You need to give all of them.

(4) Evaluate exactly sin( 1/2 Arcsin 4/5)

(5) Use a calculator to determine Arctan 10 in radians to two decimal places.

Answer 1

Answer on Question #75577 – Math – Trigonometry

Question

(1) Evaluate exactly $\tan \left(\operatorname{Arccos}\frac{2}{3}\right)$

(2) Evaluate exactly $\cos \left(\arcsin 1 + \arccos \frac{1}{2}\right)$

(3) Determine $\arcsin \frac{1}{\sqrt{2}}$. Careful: that is $\arcsin$ and not $\arcsin$, so there are going to be infinitely many answers. You need to give all of them.

(4) Evaluate exactly $\sin \left(\frac{1}{2} \arcsin \frac{4}{5}\right)$

(5) Use a calculator to determine $\arctan 10$ in radians to two decimal places.

Solution

1. Formula: $\tan(\arccos x) = \frac{\sqrt{1 - x^2}}{x}, |x| \leq 1, x \neq 0.$

$$\tan \left(\arccos \frac{2}{3}\right) = \frac{\sqrt{1 - \left(\frac{2}{3}\right)^2}}{\frac{2}{3}} = \frac{\sqrt{1 - \frac{4}{9}}}{\frac{2}{3}} = \frac{\sqrt{5}}{\frac{2}{3}} = \frac{\sqrt{5}}{3} * \frac{3}{2} = \frac{\sqrt{5}}{2}.$$

2. Formula: $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$

$$\begin{array}{l} \cos \left(\arcsin 1 + \arccos \frac{1}{2}\right) = \cos(\arcsin 1) \cos \left(\arccos \frac{1}{2}\right) - \sin(\arcsin 1) \sin \left(\arccos \frac{1}{2}\right) = \\ = \cos \frac{\pi}{2} * \frac{1}{2} - 1 * \sin \frac{\pi}{3} = 0 * \frac{1}{2} - 1 * \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \end{array}$$

3. Formula: $\arcsin a = (-1)^n \arcsin a \pm \pi n, n \in \mathbb{Z}$

$$\arcsin \frac{1}{\sqrt{2}} = (-1)^n \frac{\pi}{4} \pm \pi n, n \in \mathbb{Z}.$$

4. Formula: $\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}}; \cos(\arcsin x) = \sqrt{1 - x^2}$

$$\begin{array}{l} \sin \left(\frac{1}{2} \arcsin \frac{4}{5}\right) = \pm \sqrt{\frac{1 - \cos \left(\arcsin \frac{4}{5}\right)}{2}} = \pm \sqrt{\frac{1 - \sqrt{1 - \left(\frac{4}{5}\right)^2}}{2}} = \pm \sqrt{\frac{1 - \sqrt{\frac{9}{25}}}{2}} = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}} = \\ = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5} \end{array}$$

5. $\arctan 10 = 1.471127 \approx 1.47$ radian.

Answer: 1. $\frac{\sqrt{5}}{2}$; 2. $-\frac{\sqrt{3}}{2}$; 3. $(-1)^n \frac{\pi}{4} \pm \pi n, n \in \mathbb{Z}$; 4. $\pm \frac{\sqrt{5}}{5}$; 5. 1.47 radian.

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