Question

If (sin θ + cos θ) = a and (sin³θ + cos³θ) = b, then (3a − 2b) = ?
(a) a³
(b) b³
(c) 0
(d) 1

Answer 1

(a) a³

$$\begin{gathered} \text{Given:}(\text{sin}\theta +\text{cos}\theta )=a \\ \Rightarrow {(\text{sin}\theta +\text{cos}\theta )}^2=a^2\left[\text{Squaring both sides}\right] \\ \Rightarrow ({\text{sin}}^2\theta +{\text{cos}}^2\theta )+2\text{sin}\theta \text{cos}\theta =a^2 \\ \Rightarrow \text{1+}2\text{sin}\theta \text{cos}\theta =a^2 \\ \Rightarrow \text{sin}\theta \text{cos}\theta =\frac{a^2-1}{2} \\ \text{Again,}{(\text{sin}\theta +\text{cos}\theta )}^3=a^3\left[\mathrm{C}\text{ubing both sides}\right] \\ \Rightarrow {\text{sin}}^3\theta +{\text{cos}}^3\theta +3\text{sin}\theta \text{cos}\theta (\text{sin}\theta +\cos \theta )=a^3 \\ \therefore b+3\left(\frac{a^2-1}{2}\right)a=a^3 \\ \Rightarrow b+\frac{3a(a^2-1)}{2}=a^3 \\ \Rightarrow \text{2}b+3a(a^2-1)=2a^3 \\ \Rightarrow \text{2}b+3a^3-3a=2a^3 \\ \Rightarrow -3a+2b=-a^3 \\ \Rightarrow 3a-2b=a^3 \end{gathered}$$