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

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.

## 3 Answers

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.)

• Thank you for the effort! I really appreciate it. – Exzone Feb 17 at 18:21
• No problem! The other two were easy, but the third one was a little tougher. – bjcolby15 Feb 17 at 18:25

Hint: Compute $$\cos\left(3\arcsin\frac14+\arccos\frac{11}{16}\right)$$.

• Do I need to put any restrictions besides computing ? – Exzone Feb 17 at 17:40
• I cannot imagine which restrictions you have in mind. – José Carlos Santos Feb 17 at 17:41
• Thanks! I got it. – Exzone Feb 17 at 17:44

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

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

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