I need help to verify if this is correct and I can't find the mistake can you?
csc^2(2x)sin(x)tan(x)+2csc(x)cos(x)=sec(x)csc^2(2x )
Hello, talor!
I need help to verify if this is correct and I can't find the mistake, can you?
. . What "mistake" are you looking for?
$\displaystyle \csc^2(2x)\sin(x)\tan(x)+2\csc(x)\cos(x)\;=\;\sec( x)\csc^2(2x)$
Well, it's not an identity . . . Try $\displaystyle x = \frac{\pi}{4}$
Left side: .$\displaystyle \underbrace{\csc^2\!\left(\frac{\pi}{2}\right)}\un derbrace{\sin\left(\frac{\pi}{4}\right)}\underbrac e{\tan\left(\frac{\pi}{4}\right)} + \underbrace{2\cdot\csc\left(\frac{\pi}{4}\right)}\ underbrace{\cos\left(\frac{\pi}{4}\right)} $
. . . . . . $\displaystyle = \quad\;\; 1^2\quad\cdot\quad \frac{1}{\sqrt{2}}\quad\cdot\quad 1 \quad+\quad 2\cdot \sqrt{2} \quad\cdot\quad \frac{1}{\sqrt{2}} \quad = \quad \boxed{\frac{1}{\sqrt{2}} + 2}$
Right side: .$\displaystyle \underbrace{\sec\left(\frac{\pi}{4}\right)}\underb race{\csc^2\left(\frac{\pi}{2}\right)} $
. . . . . . $\displaystyle = \qquad \sqrt{2} \quad\cdot\quad 1^2 \quad = \quad\boxed{ \sqrt{2}}$
. . . See?
I need to prove if there are any errors and if so identify them and explain why they are errors. If there are no errors explain why. I need help get gets to confusing to follow for me. thanks erin
Csc^2(2x)sin(x)tan(x)+2 csc(x)cos(x)=sec(x)csc^2(2x)
Csc(x)[csc(x)sin(x)tan(x)+2 cos(x)]=sec(x)csc^2(2x)
[csc(x)sin(x)tan(x)+ 2 cos(x)] sec(x)csc(2x)
csc(x)sin(x)(cos(x)/sin(x)) + 2 cos(x) =sec(x)csc(x)
cos(x)[csc(x+2)] = sec(x)csc(2x)
cos(x)csc(2x)=sec(x)csc(2x)
cos(x)=sec(x)
sec(x)cos(x) = 1
(1/cos(x))cos(x) = 1
1=1
Hello, talor!
The error is that the statement is not true!
And sorry, I don't understand your steps.
. . And did you notice the $\displaystyle {\color{red}2x}$ ?
$\displaystyle \csc^2({\color{red}2x})\sin(x)\tan(x) + 2\csc(x)\cos(x)\;=\;\sec(x)\csc^2({\color{red}2x})$
The left side is: .$\displaystyle \frac{1}{\sin^2(2x)}\cdot\sin(x)\cdot\frac{\sin(x) }{\cos(x)} + 2\cdot\frac{1}{\sin(x)}\cdot\cos(x) $
. . $\displaystyle =\;\frac{\sin^2(x)}{\sin^2(2x)\cos(x)} + \frac{2\cos(x)}{\sin(x)} \;=\;\frac{\sin^2(x)}{[2\sin(x)\cos(x)]^2\cos(x)} + \frac{2\cos(x)}{\sin(x)} $
. . $\displaystyle = \;\frac{\sin^2(x)}{4\sin^2(x)\cos^3(x)} + \frac{2\cos(x)}{\sin(x)} \;=\;\frac{1}{4\cos^3(x)} + \frac{2\cos(x)}{\sin(x)} $
. . $\displaystyle = \;\boxed{\frac{\sin(x) + 8\cos^4(x)}{4\sin(x)\cos^3(x)}}\;\;{\color{blue}[1]} $
The right side is: .$\displaystyle \frac{1}{\cos(x)}\cdot\frac{1}{\sin^2(2x)} \;=\;\frac{1}{\cos(x)}\cdot\frac{1}{[2\sin(x)\cos(x)]^2} $
. . $\displaystyle = \;\frac{1}{\cos(x)}\cdot\frac{1}{4\sin^2(x)\cos^2( x)} \;=\;\boxed{\frac{1}{4\sin^2(x)\cos^3(x)}}\;\;{\co lor{blue}[2]} $
And [1] and [2] are definitely not equal . . .
I saw that and thought that maybe they were dividing the whole equation by 2 or squaring the problem but that didn't look right But I didn't know how to explain in words I knew it was wrong but don't know the right way to explain other than you can not do it. thanks for helping me. validate I knew the mistake. Erin
there are a lot of $\displaystyle x$'s for which $\displaystyle \csc (x) \not= \csc (2x)$ so you cannot just cancel it..
in fact, suppose, they are equal.
$\displaystyle \csc (x) = \csc (2x)$ can also be written as $\displaystyle \frac{1}{\sin x} = \frac{1}{\sin(2x)}$ or $\displaystyle \sin x = \sin (2x)$
then, $\displaystyle \sin x = 2\sin (x)\cos (x) \Longleftrightarrow \sin x - 2\sin (x)\cos (x) = 0 \Longleftrightarrow \sin x(1 - 2\cos (x)) = 0$
so, either $\displaystyle \sin x = 0$ or $\displaystyle 1 - 2\cos (x) = 0 \Leftrightarrow \cos x = \frac{1}{2}$
$\displaystyle \sin x = 0$ cannot happen otherwise $\displaystyle \csc x$ is undefined.
it means that $\displaystyle \cos x = \frac{1}{2}$ should be true. hence $\displaystyle x=\pm \frac{\pi}{3} + 2n\pi$ for any integer $\displaystyle n$.
this only means that $\displaystyle \csc (x) = \csc (2x)$ is true only when $\displaystyle x=\pm \frac{\pi}{3} + 2n\pi$ for any integer $\displaystyle n$..