How do you simplify $cos (arccos (2 x) + arcsin (x))$ $Cos(arccos(2x) + arcsin(x))$ ?

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How do you simplify $cos (arccos (2 x) + arcsin (x))$ $Cos(arccos(2x) + arcsin(x))$ ?

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Answer 1 · 2016-07-24T03:58:07

cos 3x

Explanation:

arccos 2x --> 2x
arcsin x --> sin x
cos (arccos 2x + arcsin x) = cos (2x + x) = cos 3x

Answer 2 · 2016-07-24T04:48:51

$\pm 2 x \sqrt{1 - x^{2}} \pm x \sqrt{1 - 4 x^{2}}, - \frac{1}{2} \leq x \leq \frac{1}{2}$ $+-2 x sqrt (1 - x^2 ) +- x sqrt(1 - 4 x^2 ), -1/2<=x<=1/2$

Explanation:

arc cos 2x is the angle whose cosine is 2x.

Let $a = a r c cos (2 x)$ $a = arc cos ( 2 x )$ .

Then, $cos a = 2 x \in [- 1, 1] and sin a = \pm \sqrt{1 - 2 x^{2}}$ $cos a = 2 x in [-1, 1 ] and sin a = +- sqrt (1 - 2 x^2 )$

Note that $x \in [- \frac{1}{2} . \frac{1}{2}]$ $x in [-1/2. 1/2]$ .

Prefix negative sign for principal value $a > \frac{π}{4}$ $a > pi/4$ (when x > 0)..

Let $b = a r c sin x$ $b = arc sin x$ .

Then, $sin b = x and cos b = \pm \sqrt{1 - x^{2}}$ $sin b = x and cos b = +- sqrt (1 - x^2 )$ .

Prefix negative sign for principal value $b < 0$ $b < 0$ (when x < 0).

Now, the given expression is

$cos (a + b)$ $cos ( a + b )$

$= cos a cos b - sin a sin b$ $= cos a cos b - sin a sin b$

$= \pm 2 x \sqrt{1 - x^{2}} \pm x \sqrt{1 - 4 x^{2}}$ $=+-2 x sqrt( 1 - x^2) +- x sqrt ( 1 - 4 x^2 )$

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How do you simplify $cos (arccos (2 x) + arcsin (x))$ $Cos(arccos(2x) + arcsin(x))$ ?

2 Answers

Explanation:

Explanation:

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How do you simplify $cos (arccos (2 x) + arcsin (x))$ $Cos(arccos(2x) + arcsin(x))$ ?