Question

find sin ⁶ tita - cos ⁶ tita

Answer 1

$$\begin{gathered} {\sin }^6\mathrm{\theta }-{\cos }^6\mathrm{\theta } \\ ={\left({\sin }^2\mathrm{\theta }\right)}^3-{\left({\cos }^2\mathrm{\theta }\right)}^3 \\ =\left({\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }\right)\left[{\sin }^4\mathrm{\theta }+{\cos }^4\mathrm{\theta }+{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right]\left[\mathrm{As}\mathrm{we}\mathrm{know}\mathrm{that}{\mathrm{x}}^3-{\mathrm{y}}^3=\left(\mathrm{x}-\mathrm{y}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}\right)\right] \\ =\left({\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }\right)\left[{\left({\sin }^2\mathrm{\theta }\right)}^2+{\left({\cos }^2\mathrm{\theta }\right)}^2+2{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }-{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right] \\ =\left({\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }\right)\left[{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)}^2-{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right] \\ =\left({\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }\right)\left[{\left(1\right)}^2-{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right] \\ =\left({\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }\right)\left[1-{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right] \end{gathered}$$