How can I prove this equation is an identity? sin^4w=1-cot^2w + cos^2w (cot^2w)/(csc^2w)

Mar 25, 2018

This is not an identity.

Explanation:

The left hand side is

${\sin}^{4} w = {\left(1 - {\cos}^{2} w\right)}^{2} = 1 - \textcolor{red}{2 {\cos}^{2} w} + {\cos}^{4} w$

while the right hand side simplifies to

$1 - {\cot}^{2} w + {\cos}^{2} w \frac{{\cot}^{2} w}{{\csc}^{2} w} = 1 - {\cot}^{2} w + {\cos}^{2} w \frac{{\cos}^{2} w}{{\sin}^{2} w {\csc}^{2} w} = 1 - \textcolor{red}{{\cot}^{2} w} + {\cos}^{4} w$

Since ${\cot}^{2} w \ne 2 {\cos}^{2} w$ in general, this is not an identity.

Mar 25, 2018

it is not an identity

Explanation:

We seek to prove:

${\sin}^{4} w \equiv 1 - {\cot}^{2} w + {\cos}^{2} w \left({\cot}^{2} \frac{w}{\csc} ^ 2 w\right)$

We can readily disprove the claim using a counter example:

Consider the case $w = \frac{\pi}{6}$, then:

$L H S = {\sin}^{4} \left(\frac{\pi}{6}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\left(\frac{1}{2}\right)}^{4}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{16}$

And:

$R H S = 1 - {\cot}^{2} \left(\frac{\pi}{6}\right) + {\cos}^{2} \left(\frac{\pi}{6}\right) {\cot}^{2} \frac{\frac{\pi}{6}}{\csc} ^ 2 \left(\frac{\pi}{6}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 - {\left(\sqrt{3}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2} {\left(\sqrt{3}\right)}^{2} / {\left(2\right)}^{2}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 - 3 + \left(\frac{3}{4}\right) \frac{3}{4}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 2 + \frac{9}{16}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{23}{16}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \ne L H S$

Hence this is not an identity