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How would I solve the following trig equation?

$\arctan(x)+\arcsin(x)=\frac \pi 2$

I am kind of confused on how to solve it.

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We can obviously rearrange it to $$\arcsin(x)=\frac{\pi}{2}-\arctan(x),$$, which is equivalent to $$x=\sin(\frac{\pi}{2}-\arctan(x)).$$ Using the difference-angle formula, we have $$x=\cos(\arctan(x)).$$ Now, viewing $x=\frac{x}{1}$, we can think of $\cos(\arctan(x))$ as the sides of a triangle, in particular, we have $$x=\frac{1}{\sqrt{x^2+1}}.$$ Squaring and multiplying both sides by $x^2+1$ quickly yields the equation $$x^4+x^2-1=0.$$ Set $w=x^2$ so that the equation above becomes $$w^2+w-1=0.$$ The quadratic formula gives us $$x^2=w=\frac{-1\pm\sqrt{1-4(1)(-1)}}{2(1)}.$$ Simplifying the expression and observing that $x^2\geq0$ for all real numbers $x$, we have $$x=\sqrt{\frac{-1+\sqrt{5}}{2}}.$$ I hope this helps!

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    One quick question I have is how did you know x=sin((pi/2-arctan(x)) x=cos(arctan(x)2012-12-16
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    The difference angle formula for $\sin(\alpha-\beta)$ is $\sin(\alpha)\cos(\beta)-\sin(\beta)\cos(\beta)$. See if you can verify the work from there.2012-12-16
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    I am still confused because sin(pi/2)=1 cos(pi/2)=0 arctan(x) tan=(x/1) so sin would be (x)/square root(x+1) and cos would be 1/(square root(x+1).2012-12-16
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    With $\arctan(x)$, the legs of the right triangle will be $x$ and $1$ where $x$ is the side opposite your angle, and the leg of length $1$ is adjacent to the angle. Using the Pythagorean Thm, we have the hypotenuse is $\sqrt{x^2+1}$. Checking that cosine is the adjacent leg over the hypotenuse, we have $$\cos\big(\arctan(x)\big)=\frac{1}{\sqrt{x^2+1}}.$$ Does this help at all?2012-12-16
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    I understand how you got the part you described, I am having trouble getting the part when you get x=sin(pi/2-arctan(x) and then got cos(arctan(x)2012-12-16
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    I see. Well, we know $\arcsin(x)=\frac{\pi}{2}-\arctan(x)$. Note that $\sin\big(\arcsin(x)\big)=x$, so all I had done is apply the sine function to both sides. After applying sine to both sides, I used the difference-angle formula to reduce it to the above.2012-12-16
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    Hmm I guess it is difficult to explain I understand how you got (1/square root(x^2+1) , my question is how did you use the difference angle formula to go from x=sin((pi/2)-arctan(x)) to cos(arctan(x)).2012-12-16
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    $\sin(\frac{\pi}{2}-\arctan(x))=\sin(\frac{\pi}{2})\cos(\arctan(x))-\sin(\arctan(x))\cos(\frac{\pi}{2})$. As you noted above, $\sin(\frac{\pi}{2})=1$ and $\cos(\frac{\pi}{2})=0$, so the second term drops out, while the first one reduces to $\cos(\arctan(x))$.2012-12-16
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    Oh thanks I see now it makes sense.2012-12-16
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    I have one more question to ask that has been troubling me.When you do (x^2+1)(x^2+1) would you not get x^4+2x^2+2 Maybe I a not seeing something.2012-12-16
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    The square root and the square "cancel" each other.2012-12-16
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    But if the square root cancel the square you get (x+1) and then if you multiply times (x^2+1) you have x^3+x^2+2x2012-12-16
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    $$x^2=(x)^2=\left(\frac{1}{\sqrt{x^2+1}}\right)^2=\frac{1}{x^2+1}$$ Now see if you can multiply and get what I have above.2012-12-16
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    Sorry to ask so many questions I got (x^2+1)(x^2+1)=(x^4+2x^2+1)2012-12-16
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    @FernandoMartinez: We have $x^2=\frac{1}{x^2+1}$. Multiply both sides by $x^2+1$. We get $x^2(x^2+1)=1$. This simplifies to $x^4+x^2=1$, and then to $x^4+x^2-1=0$.2012-12-22