Question

Choose the correct option:
If $sinA+cosA=x$ and $si{n}^{3}A+co{s}^{3}A=y$, then:

• A. $x^3+2x-3y=0$
• B. $x^3+3x+2y=0$
• C. $x^3-3x+2y=0$
• D. $x^3-3x-2y=0$

Answer 1

The correct option is C $x^3-3x+2y=0$

$y=(sinA{)}^3+(cosA{)}^3\Rightarrow y=(sinA+cosA)(si{n}^{2}A+co{s}^{2}A-sinAcosA)[using{a}^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]\Rightarrow y=x(si{n}^{2}A+co{s}^{2}A+2sinAcosA-3sinAcosA)\Rightarrow y=x\left[\right(sinA+cosA{)}^2-3sinAcosA]\Rightarrow y=x[x^2-3sinAcosA]\to (1)$
Now, $x=sinA+cosA\Rightarrow x^2=(si{n}^{2}A+co{s}^{2}A)+2sinAcosA\Rightarrow \frac{x^2-1}{2}=sinAcosA\to \left(2\right)$
$\therefore$ from (1)
$y=x[x^2-\frac{3}{2}(x^2-1\left)\right]\Rightarrow y=x(\frac{3}{2}-\frac{x^2}{2})\Rightarrow 2y=3x-2\Rightarrow x^3-3x+2y=0$