i was just practicing my trigonometry. And i always find finding value of trigonometric inverse functions without a calculator to be hard. Can you guys give me questions to work with? For example, finding $\theta$ such that $$\theta = \arctan(2-\sqrt3)$$
Thanks in advance. Don't go easy on me :)
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SchrodingersCat
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user283172
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1Find $\arctan1+\arctan2+\arctan3$. – Ivan Neretin Nov 03 '15 at 13:32
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It is hard,and it almost not meant to be done by hand,almost never. – TheCoolDrop Nov 03 '15 at 13:32
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To name just a few..
- Evaluate $\sin(\arctan(4))$
- Find the inverse of $1 + 2\cos\left(\frac{\pi}{3(t+1)}\right)$
- Prove $\arctan(a) + \arctan(b) = \begin{cases} \arctan \left(\frac{a+b}{1-ab}\right), & ab < 1 \\ \arctan \left(\frac{a+b}{1-ab}\right) + \pi, & ab > 1, a,b > 0 \\ \arctan \left(\frac{a+b}{1-ab}\right) - \pi, & ab > 1, a,b < 0 \end{cases}$
- Find $\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right)$
More to be added, stay tuned!
