Question

Is LHS=RHS?
${\sin }^8\theta -{\cos }^8\theta =({\sin }^2\theta -{\cos }^2\theta )(1-2{\sin }^2\theta {\cos }^2\theta )$

• A. Yes
• B. No
• C. Ambiguous
• D. Can't say

Answer 1

Answer: (A)

Yes

$si{n}^8\theta -co{s}^8\theta$
$=(si{n}^4\theta -co{s}^4\theta )(si{n}^4\theta +co{s}^4\theta )$
$=(si{n}^2\theta -co{s}^2\theta )(si{n}^2\theta +co{s}^2\theta )\left(\right(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta .co{s}^2\theta )$
$=(si{n}^2\theta -co{s}^2\theta )(1-2si{n}^2\theta .co{s}^2\theta )$