Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 156795 by tabata last updated on 15/Oct/21

$$prove\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(sinx\right)}^{2n}={sec}^{2}x???$$

Answered by FongXD last updated on 15/Oct/21

$$\mathrm{we}\mathrm{have}:\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left(\mathrm{sinx}\right)}^{2\mathrm{n}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left({\sin }^2\mathrm{x}\right)}^{\mathrm{n}}$$$$\mathrm{since}-1⩽\mathrm{sinx}⩽1,\Rightarrow 0⩽{\sin }^2\mathrm{x}<1,\mathrm{\forall }\mathrm{x}\in \mathbb{R}-\{\frac{\pi }{2}(4\mathrm{k}\pm 1),\mathrm{k}\in \mathbb{Z}\}$$$$\mathrm{therefore},\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left({\sin }^2\mathrm{x}\right)}^{\mathrm{n}}\mathrm{is}\mathrm{the}\mathrm{infinite}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{geometric}\mathrm{sequence}$$$$\mathrm{whose}\mathrm{first}\mathrm{term}\mathrm{is}{\left({\sin }^2\mathrm{x}\right)}^0=1\mathrm{and}\mathrm{common}\mathrm{ratio}\mathrm{r}={\sin }^2\mathrm{x}$$$$\mathrm{Formula}:{\mathrm{S}}_{\mathrm{\infty }}=\frac{{\mathrm{u}}_1}{1-\mathrm{q}}$$$$\mathrm{we}\mathrm{get}:\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left({\sin }^2\mathrm{x}\right)}^{\mathrm{n}}=\frac{1}{1-{\sin }^2\mathrm{x}}=\frac{1}{{\cos }^2\mathrm{x}}={\sec }^2\mathrm{x}✓$$$$\mathrm{therefore}.\overline{)\begin{array}{c}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left(\mathrm{sinx}\right)}^{2\mathrm{n}}={\sec }^2\mathrm{x}\mathrm{is}\mathrm{indeed}\mathrm{true}\end{array}}$$

Commented by Ruuudiy last updated on 17/Oct/21

$$\mathrm{Thanks}\mathbf{Ser}.\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\sin x\right)}^{2n}={\sec }^{2}x\mathrm{is}\mathrm{indeed}\mathrm{true}.$$

Answered by som(math1967) last updated on 15/Oct/21

$$1+{sin}^{2}x+{sin}^{4}x+...\mathrm{\infty }$$$$=\frac{1}{1-{sin}^{2}x}=\frac{1}{{cos}^{2}x}={sec}^{2}x$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com