Question

If $\sin \left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right)+{\sin }^3\left(\mathrm{x}\right)=1$ then find the value of ${\cos }^6\left(\mathrm{x}\right)-4{\cos }^4\left(\mathrm{x}\right)+8{\cos }^2\left(\mathrm{x}\right)$.

Answer 1

Use basic trigonometric identities

Given, $\sin \left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right)+{\sin }^3\left(\mathrm{x}\right)=1$

$$\begin{gathered} \begin{array}{ccc}& {\sin }^3\left(\mathrm{x}\right)+\sin \left(\mathrm{x}\right)=1-{\sin }^2\left(\mathrm{x}\right)& \\ \Rightarrow & \sin \left(\mathrm{x}\right)\left({\sin }^2\left(\mathrm{x}\right)+1\right)={\cos }^2\left(\mathrm{x}\right)& \left[\because {\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)=1\right]\end{array} \\ \Rightarrow \sin \left(x\right)(1-{\cos }^2(x)+1)={\cos }^2\left(x\right) \\ \Rightarrow \sin \left(x\right)(2-{\cos }^2(x\left)\right)={\cos }^2\left(x\right) \end{gathered}$$

Squaring on both sides

$\begin{array}{ccc}\Rightarrow & {\sin }^2\left(\mathrm{x}\right)\left[4+{\cos }^4\left(\mathrm{x}\right)-4{\cos }^2\left(\mathrm{x}\right)\right]={\cos }^4\left(\mathrm{x}\right)& \left[\because {\left(\mathrm{a}-\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\right]\\ \Rightarrow & \left[1-{\cos }^2\left(\mathrm{x}\right)\right]\left[4+{\cos }^4\left(\mathrm{x}\right)-4{\cos }^2\left(\mathrm{x}\right)\right]={\cos }^4\left(\mathrm{x}\right)& \left[\because {\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)=1\right]\\ \Rightarrow & 4+5{\cos }^4\left(\mathrm{x}\right)-8{\cos }^2\left(\mathrm{x}\right)-{\cos }^6\left(\mathrm{x}\right)={\cos }^4\left(\mathrm{x}\right)& \\ \Rightarrow & {\cos }^6\left(\mathrm{x}\right)-4{\cos }^4\left(\mathrm{x}\right)+8{\cos }^2\left(\mathrm{x}\right)=4& \end{array}$

Hence, value of ${\cos }^6\left(\mathrm{x}\right)-4{\cos }^4\left(\mathrm{x}\right)+8{\cos }^2\left(\mathrm{x}\right)$ is $4$.