Question

Hi
How do we solve the following: (using an identity!! the answer is : rad 3 over 2 )
arcsin x - arccos x = pi/6

Answer 1

Wait. What's the question again? :)

Answer 2

arcsin x - arccos x = pi/2

Answer 3

Ooops I meant it equals pi/6

Answer 4

\[\arcsin(x)=\arccos(x)+\frac{ \pi}{6}\]
\[x=\sin(\arccos(x)+\frac{ \pi}{6})\]
\[x=\sin(\arccos(x)) \cos(\frac{\pi}{6})+\sin(\frac{\pi}{6}) \cos(\arccos(x))\]
So to write sin(arccos(x)) as an algebraic expression
we need to draw a right triangle and use arccos(x)=y => cos(y)=x

so \[\sin(\arccos(x))=\sqrt{1-x^2}\]
So we have
\[x=\sqrt{1-x^2} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot x\]
Can you solve this algebraic expression for x?
\[x-\frac{1}{2}x=\frac{\sqrt{3}}{2}\sqrt{1-x^2}\]
\[\frac{1}{2}{x}=\frac{\sqrt{3}}{2}\sqrt{1-x^2}\]
Square both sides
\[\frac{1}{4}x^2=\frac{3}{4}(1-x^2)\]
\[\frac{1}{4}x^2=\frac{3}{4}-\frac{3}{4}x^2\]

I will leave the rest to you :)