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Question Number 196639 by bbbbbbbb last updated on 29/Aug/23

$$\mathrm{if}\mathrm{sinx}=\frac{2}{3}\mathbf{then}\mathbf{find}\mathbf{you}\mathbf{the}\mathbf{value}\mathbf{of}{\mathbf{sin}}^6\mathbf{x}+{\mathbf{cos}}^6\mathbf{x}???$$

Answered by deleteduser1 last updated on 28/Aug/23

$$={({sin}^{2}x+{cos}^{2}x)}^3-3{sin}^2{cos}^{2}x({sin}^{2}x+{cos}^{2}x)$$$$=1-3{sin}^2\left(x\right)(1-{sin}^{2}x)\left(1\right)=1-3\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)=\frac{7}{27}$$

Answered by Frix last updated on 28/Aug/23

$$s=\frac{2}{3}$$$$c=\pm \sqrt{1-s^2}=\pm \frac{\sqrt{5}}{3}$$$$s^6+c^6=\frac{7}{27}$$$$$$$$s^6+c^6=s^6+{(1-s^2)}^3=3s^4-3s^2+1=\frac{7}{27}$$

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