How do you write the trigonometric expression $cos (arccos x + arcsin x)$ $cos(arccosx+arcsinx)$ as an algebraic expression?

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How do you write the trigonometric expression $cos (arccos x + arcsin x)$ $cos(arccosx+arcsinx)$ as an algebraic expression?

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Answer 1 · 2018-01-20T13:42:09

The answer is $= 0$ $=0$

Explanation:

Let $y = arccos x$ $y=arccosx$ , $\Rightarrow$ $=>$ , $cos y = x$ $cosy=x$

Let $z = arcsin x$ $z=arcsinx$ , $\Rightarrow$ $=>$ , $sin z = x$ $sinz=x$

Therefore,

$cos (arccos x + arcsin x) = cos (y + z)$ $cos(arccosx+arcsinx)=cos(y+z)$

$= cos y cos z - sin y sin z$ $=cosycosz-sinysinz$

$= x \cdot \sqrt{1 - x^{2}} - x \sqrt{1 - x^{2}}$ $=x*sqrt(1-x^2)-xsqrt(1-x^2)$

$= 0$ $=0$

Answer 2 · 2018-01-20T13:51:37

$\Rightarrow cos (arccos x + arcsin x) = 0$ $=> cos( color(red)(arccosx) + color(blue)(arcsinx )) = 0$

$\forall | x | \leq 1$ $AA |x| <=1$

$for all x that is in the interval - 1 \leq x \leq 1$ $" for all " x " that is in the interval " -1<= x <= 1$

Explanation:

The first thing we need to consider is our additional formulae for trig:

$cos (A + B) = cos A cos B - sin A sin B$ $cos(A + B ) = cosAcosB - sinAsinB$

So for this problem we see that $A = arccos x$ $color(red)(A = arccosx$ and $B = arcsin x$ $color(blue)(B = arcsinx$

$\Rightarrow cos (arccos x + arcsin x) =$ $=> cos( color(red)(arccosx) + color(blue)(arcsinx )) =$

$cos (arccos x) \cdot cos (arcsin x) - sin (arccos x) \cdot sin (arcsin x)$ $cos( color(red)(arccosx) ) * cos color(blue)((arcsinx )) - sin color(red)( (arccosx) ) * sin color(blue)( (arcsinx) )$

We know:

$cos (arccos x) = x and sin (arcsin x) = x$ $cos(arccosx) = x " and " sin(arcsinx) = x$

Now to find $cos (arcsin x) and sin (arccos x) :$ $cos(arcsinx) " and " sin(arccosx ) :$

$let θ_{1} = arcsin x$ $"let " color(orange)(theta_1 = arcsinx$

$\Rightarrow {sin θ}_{1} = x$ $=> sin theta_1 = x$

Use ${cos}^{2} x + {sin}^{2} \equiv 1$ $color(green)( cos^2 x + sin^2 -= 1$

${cos θ}_{1} = \sqrt{1 - {sin}^{2} θ_{1}} = \sqrt{1 - x^{2}}$ $costheta_1 = sqrt(1-sin^2 theta_1 ) = sqrt(1-x^2 )$

$\Rightarrow cos (θ_{1}) = cos (arcsin x) = \sqrt{1 - x^{2}}$ $=> cos(theta_1) = cos(arcsinx ) = sqrt(1-x^2 )$

$let θ_{2} = arccos x$ $"let " color(orange)(theta_2 = arccosx$

$\Rightarrow {cos θ}_{2} = x$ $=> costheta_2 = x$

$\Rightarrow {sin θ}_{2} = \sqrt{1 - x^{2}}$ $=> sintheta_2 = sqrt(1-x^2 )$

$\Rightarrow sin (arccos x) = \sqrt{1 - x^{2}}$ $=> sin(arccosx) = sqrt(1-x^2 )$

Hence this:

$cos (arccos x) \cdot cos (arcsin x) - sin (arccos x) \cdot sin (arcsin x)$ $cos( color(red)(arccosx) ) * cos color(blue)((arcsinx )) - sin color(red)( (arccosx) ) * sin color(blue)( (arcsinx) )$

Becomes:

$x \sqrt{1 - x^{2}} - x \sqrt{1 - x^{2}} = 0$ $xsqrt(1-x^2) - xsqrt(1-x^2) = color(red)(0$

Note: This only holds for $| x | \leq 1$ $|x| <= 1$

As that is values of $x$ $x$ that $arcsin x$ $arcsinx$ and $arccos x$ $arccosx$ are valid for...

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How do you write the trigonometric expression $cos (arccos x + arcsin x)$ $cos(arccosx+arcsinx)$ as an algebraic expression?

2 Answers

Explanation:

Explanation:

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How do you write the trigonometric expression cos(arccosx+arcsinx)cos(arccosx+arcsinx) as an algebraic expression?

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How do you write the trigonometric expression $cos (arccos x + arcsin x)$ $cos(arccosx+arcsinx)$ as an algebraic expression?