Question

If $a=sin\theta +\cos \theta ,b=si{n}^3\theta +{\cos }^3\theta$ then

• A. $a^3-3a+2b=0$
• B. $a^3+3a+2b=0$
• C. $a^3-3a-2b=0$
• D. $a^3+3a-2b=0$

Answer 1

The correct option is A $a^3-3a+2b=0$

$a=sin\theta +cos\theta$
$b={sin}^3\theta +{cos}^3\theta$
$a^3={sin}^3\theta +{cos}^3\theta +3sin\theta cos\theta (sin\theta +cos\theta )$
$3a=3(sin\theta +cos\theta )$
$a^3-3a={sin}^3\theta +{cos}^3\theta -3(sin\theta +cos\theta )(sin\theta cos\theta -1)$
$={sin}^3\theta +{cos}^3\theta -3({sin}^3\theta +{cos}^3\theta )$
$=-2({sin}^3\theta +{cos}^3\theta )$
$=-2b$
Or, $a^3-3a+2b=0$
Hence, option 'A' is correct.