# How to prove sin^6x+cos^6x=1-3sin^2xcos^2x?

May 23, 2018

$L H S = {\sin}^{6} x + {\cos}^{6} x$

$= {\left({\sin}^{2} x\right)}^{3} + {\left({\cos}^{2} x\right)}^{3}$

Using formula

${a}^{3} + {b}^{3} = {\left(a + b\right)}^{3} - 3 a b \left(a + b\right)$

$= {\left({\sin}^{2} x + {\cos}^{2} x\right)}^{3} + 3 {\sin}^{2} x {\cos}^{2} x \left({\sin}^{2} x + {\cos}^{2} x\right)$

$= {1}^{3} + 3 {\sin}^{2} x {\cos}^{2} x \cdot 1$

$= 1 - 3 {\sin}^{2} x {\cos}^{2} x = R H S$

May 23, 2018

See below.

#### Explanation:

Proof

${\sin}^{6} \left(x\right) + {\cos}^{6} \left(x\right)$

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

$= {\sin}^{4} \left(x\right) - {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right) + {\cos}^{4} \left(x\right)$ (2)

$= \left(1 - 2 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right)\right) - {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right)$ (3)

$= 1 - 3 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right)$

More detailed explanations

(1) since ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$,

(2) since ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$,

(3) since ${\left({\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right)}^{2} = 1 \iff {\sin}^{4} \left(x\right) + {\cos}^{4} \left(x\right) + 2 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right) \iff {\sin}^{4} \left(x\right) + {\cos}^{4} \left(x\right) = 1 - 2 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right)$

May 23, 2018

Call ${\sin}^{2} x = u$, and ${\cos}^{2} x = v$
We get, knowing that (u + v = 1):
${u}^{3} + {v}^{3} = \left(u + v\right) \left({u}^{2} - u v + {v}^{2}\right) = {u}^{2} + {v}^{2} - u v \left(1\right)$
Note that:
${u}^{2} + {v}^{2} = {\left(u + v\right)}^{2} - 2 u v = 1 - 2 u v$
Equation (1) becomes:
${u}^{3} + {v}^{3} = 1 - 2 u v - u v = 1 - 3 u v$

${\sin}^{6} x + {\cos}^{6} x = 1 - 3 {\sin}^{2} x . {\cos}^{2} x$