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Can someone help me with this exercise? I honestly don't know where to start and how to prove it. You don't have to answer it fully, just give me a hint or something. Thank you in advance.

Exercise 1. Prove that $3\arcsin \frac{1}{4} + \arccos \frac {11}{16} = \frac {\pi}{2}$

Thanks.

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Hints for all three equations: exploit the complementary function of the angle, i.e. if you have $\sin \theta$, $\cos \theta = \sin (\pi/2 - \theta)$, and $\tan \theta = \dfrac {1}{\tan (\pi/2 - \theta)}.$

Hint for #1: The identity $$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$$ will help if you use another variable (say, $\alpha$) for one of the values (or simply apply this without using another variable)

Hints for #2: Use the identity $$\tan 2 \theta = \dfrac {2 \tan \theta}{1-\tan^2 \theta}$$

Hints for #3: This time, let one of the angles be $\alpha$ and the other $\beta$. Then use the identity $$\cos (2\alpha + \beta) = \cos 2 \alpha \cos \beta - \sin 2 \alpha \sin \beta$$ (suggestion: expand $\cos 2\alpha$ and $\sin 2\alpha$ also - the radicals look beastly but they do work out.)

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  • $\begingroup$ Thank you for the effort! I really appreciate it. $\endgroup$ – Exzone Feb 17 at 18:21
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    $\begingroup$ No problem! The other two were easy, but the third one was a little tougher. $\endgroup$ – bjcolby15 Feb 17 at 18:25
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Hint: Compute $\cos\left(3\arcsin\frac14+\arccos\frac{11}{16}\right)$.

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    $\begingroup$ Do I need to put any restrictions besides computing ? $\endgroup$ – Exzone Feb 17 at 17:40
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    $\begingroup$ I cannot imagine which restrictions you have in mind. $\endgroup$ – José Carlos Santos Feb 17 at 17:41
  • $\begingroup$ Thanks! I got it. $\endgroup$ – Exzone Feb 17 at 17:44
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Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,

as $0<\arcsin\dfrac14<\arcsin\dfrac12=?$ (see https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values)

$$3\arcsin\dfrac14=\arcsin\left(\dfrac34-4(1/4)^3\right)=\arcsin\dfrac{12-1}{16}$$

Finally use Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$

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  • $\begingroup$ Thanks for the effort as this was a stupid question by me. $\endgroup$ – Exzone Feb 17 at 18:13
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    $\begingroup$ @Exzone, No question is stupid if it is followed by sincere effort. Hope the links will clear the concept $\endgroup$ – lab bhattacharjee Feb 17 at 18:15
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    $\begingroup$ I appreciate that! Thanks . Though I got few down votes, don't know why... $\endgroup$ – Exzone Feb 17 at 18:16

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