1
$\begingroup$

Solve:

$$\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$$

I think the algebraic solution should start like:

$$\arcsin(x) = \frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)$$ $$x = \sin\left(\frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)\right)$$

at that stage probably I should use some trigonometric relation or property of $\arccos(x)$ to convert it to $\arcsin(x)$, however I can't figure it out.

As for the solution by graph I can't even think what steps should I follow to build it.

$\endgroup$
2
$\begingroup$

$$\arcsin x + \arccos\frac{x}{2} = \frac{5\pi}{6}$$

$$\arcsin x = \frac{5\pi}{6} - \arccos\frac{x}{2}$$ $$x = \sin{\left(\frac{5\pi}{6} - \arccos\frac{x}{2}\right)}$$ $$x=\sin\frac{5\pi}6\cos \left(\arccos\frac{x}{2}\right)-\cos\frac{5\pi}6 \sin\left(\arccos\frac{x}{2}\right)$$ $$x=\frac12\cdot\frac x2+\frac{\sqrt3}{2} \cdot\sqrt{1-\left(\frac x2\right)^2}$$ Answer: $x=1$

$\endgroup$
1
$\begingroup$

First you have the sine of one thing minus another. $$ \sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta. $$ After applying that, you will have $$ \sin\arccos w = \sqrt{1-w^2} \tag 1 $$ $$ \cos\arccos w = w $$ To get the identity $(1)$, just draw a triangle. If $\cos\alpha =w$ then you have $\text{adjacent} = w$ and $\text{hypotenuse} = 1$, so by the Pythagorean theorem you have $\text{opposite} = \sqrt{1-w^2}.$ Then recall that $\sin=\dfrac{\text{opposite}}{\text{hypotenuse}}.$

$\endgroup$
1
$\begingroup$

HINT: $$\arccos\dfrac x2-\dfrac\pi3=\dfrac\pi2-\arcsin x=\arccos x$$

Applying cosine $$\dfrac x2\cdot\cos\dfrac\pi3+\sin\dfrac\pi3\sqrt{1-\left(\dfrac x2\right)^2}=x$$

$$\implies x=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.