Find the value of $\sqrt{\left(\cos\theta\right)^2}$ at $\theta=\pi$   There can be two ways in which people will do this:   First way: Put the value of theta directly, so $\sqrt{\left(\cos\pi\right)^2}$, answer will be $1$ Second way: Simplify the expression first and then put the value of $\theta$, so in this case one will get $-1$.      Why do these two ways give different answers?
34782
There is a formula given in my module:  $$ \sqrt[n]{a^n} = \begin{cases}  \, a  &amp;\text{ if $n$ is odd } \\  |a| &amp;\text{ if $n$ is even } \end{cases} $$  I don't really understand the differences between them, kindly explain with an example.
102490
The question  is to find $x\in\left(0,\frac{\pi}{2}\right)$:  $$\frac {\sqrt3 - 1}{\sin x}  + \frac {\sqrt3  + 1}{\cos x} = 4\sqrt2 $$  What I did was to take the $\cos x$  fraction    to the right and try to simplify ;  But it looked very messy and trying to write $\sin x$  in terms of $\cos x$ didn't help.   Is there a more simple (elegant) way to do this.
3981
As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
119613
Question:  $$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$  What we did:  we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the second method we got to this: $\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1}  $, for which we know how to solve the first and second integral but not the third...  Thanks
80166
I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.  $w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$  Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.  But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?  And can one just 'let $z^2=w$' as above?  Edit: $w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
55059
My answer for this question was $2\sin(θ - 30)$, but the answer in the textbook says $-2\sin(θ - 30)$. I tried to use the concept of odd functions so using the idea that $-\sin(θ)= \sin(-θ)$.   So $-2\sin(θ-30) = 2\sin(30-θ)$ but that still is not the same as my answer and I was wondering why?   In addition, the question also asks to give the maximum and minimum values of the function and the values between $0$ and $360$ at which they occur.  I get:   max value $y = 2$ max $x$ at $θ = 120$  min $y$ value = $ -2$ min $x$ at $θ = 300$  But the textbook answers say:   max value $y = 2$ at $θ = 300$  min value $y = -2$ at $θ = 106.3$   I am quite confused as to why.  Thank you.
117251
I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!
102553
$$\tan\left(\frac{\pi}{2} -\theta \right) =  \cot \theta$$  I can prove this by changing into $\cos$ and $\sin$  But I want to know if it is possible to prove it using relation given below. If not why ? $$\tan\left(\frac{\pi}{2} -\theta \right) =\frac{ \tan\frac{\pi}{2} -\tan\theta}{1+ \tan\frac{\pi}{2}\tan\theta} $$
123137
Simplify: $x^2 &gt; 1$.  My solution: Taking square root on both sides:  $±x &gt; ±1$  So my results are:   $x &gt; 1$ $x &gt; -1$ $-x &gt; 1$ $\implies$ $(-1 &gt; x)$ $-x &gt; -1$ $\implies$ $(1 &gt; x) $   But I strongly feel this is wrong. What is wrong here?  A step-by-step explanation will help me.
99380
The real numbers $p$, $q$, $r$, $s$ satisfy    $$p^2+q^2+r^2+1=s+\sqrt{p+q+r-s}$$    Find the value of $s$.   I don't even know how to start. Thanks!
53108
In a problem, we have to differentiate : $$ y = \arcsin (2 x \sqrt{1 - x^2}),$$ given: $$-{1/(\sqrt2)} &lt; x &lt;{1/(\sqrt2)}$$.   My approach :  Put $x = \cos (p)$, so y reduces to:- $y = 2p = 2 \arccos (x)  \implies$ ${dy}/{dx}$ $ = -{2}/{\sqrt{1-x^2}}$   But the solution given is without the minus sign, so I am doing something wrong here but don't know what.  So please tell me what I am doing wrong and how to correct it.
34440
I'm having a trouble with this integral expression:  $$\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}$$  I've done this substitution: $t= \tan(\theta/2)$  and get: $\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2}{1+t^2}dt$ where $\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}$  then the integral becomes:  $$\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)$$   However, I'm not sure about the new limits since $\tan$ has period $\pi$ so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know!
18661
How do we get limit such as $0\le\theta\le\pi$, $0\le\phi\le2\pi$ in spherical coordinate system where $$x=r \sin\theta\cos\phi, y=r \sin\theta\sin\phi, z=r \cos\theta$$  Why is the $\theta$-limit $[0,\pi]$ and not $2 \pi$?  I think that the angle of the foot of perpendicular from the point taken in spherical coordinate can lie within $[0,2 \pi]$
41748
Possible Duplicate:          The square root of a number squared is equal to the absolute value of that number. Why is $\sqrt{x^2} = |x|$? Why not just $x$? Please give me a reason and also help me prove it.
116573
Suppose that you have to prove the trig identity:  $$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta$$  I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do:  $$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}$$ $$=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}$$ $$=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$ $$=\sin\theta$$  And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.  I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this:  $$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta$$ $$\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)$$ $$\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)$$ $$(\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)$$  Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true?  Or, why can't I take the identity equation, manipulate it, arrive at $(\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)$, and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct?  Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step  $$\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$ $$=\sin\theta$$  is not valid when theta is $\pi/2$, for example, because then it constitutes division by zero.
105420
If $A + B + C = \pi$, then find the minimum value of $\cot^2 A +\cot^2 B + \cot^2 C$.   I don't know how to solve it. And can you please mention the used formulas first.  What I can see is that if one of the angles $A$, $B$, $C$ is small, then the value $\cot^2A$ or $\cot^2B$ or $\cot^2C$ will be big. So I want to make angles big (more precisely, close to $\pi/2$, where cotangent is zero), but the condition $A+B+C=\pi$ prevents me from making all three of them very big .  I can see that if $A=B=C=\frac\pi3$, then I get $\cot A=\cot B=\cot C=\frac1{\sqrt3}$ and $\cot^2A+\cot^2B+\cot^2C=1$. But I do not know whether this is indeed minimum. (According to WolframAlpha . However, I would like to see some proof of this fact.)
104018
If we have to integrate $$f(x)=\frac{x}{\sqrt{1-x^2}}$$ and we substitute $x=\sin \theta$ then we eventually have to take the square root of $\cos^2x$ which is equal to $|\cos x|$.  But in my textbook and class lectures, we simply remove the absolute value sign and replace it with $\cos x$.  Why do we do this? I don't understand how we can do this.