If √2√1+2 = 17 and 1-22 = 1√10, ,(0,2), then (+2) is equal to ___________ Answer: 1

Question

If $\dfrac{\sqrt2\sin\alpha}{\sqrt{1+\cos2\alpha}} = \dfrac17$ and $\sqrt{\dfrac{1-\cos2\beta}2} = \dfrac1{\sqrt{10}}$, $\alpha,\beta\in\left(0,\dfrac\pi2ight)$, then $	an(\alpha+2\beta)$ is equal to \_\_\_\_\_\_\_\_\_\_\_

Answer: 1

Accepted Answer

Since $\sqrt{1+\cos2\alpha}=\sqrt2\cos\alpha$: $\dfrac{\sqrt2\sin\alpha}{\sqrt2\cos\alpha}=\tan\alpha=\dfrac17$.

Since $\sqrt{\dfrac{1-\cos2\beta}2}=\sin\beta$: $\sin\beta=\dfrac1{\sqrt{10}}$, so $\cos\beta=\dfrac3{\sqrt{10}}$, $\tan\beta=\dfrac13$.

$$\tan2\beta = \frac{2\tan\beta}{1-\tan^2\beta} = \frac{2/3}{1-1/9} = \frac{2/3}{8/9} = \frac34$$

$$\tan(\alpha+2\beta) = \frac{\tan\alpha+\tan2\beta}{1-\tan\alpha\tan2\beta} = \frac{\tfrac17+\tfrac34}{1-\tfrac17\cdot\tfrac34} = \frac{4/28+21/28}{1-3/28} = \frac{25/28}{25/28} = 1$$

Solutions · 1