Solve: ${\mathrm{cos}}^{2}x>{\mathrm{sin}}^{2}x$

Lydia Carey
2022-06-24
Answered

Solve: ${\mathrm{cos}}^{2}x>{\mathrm{sin}}^{2}x$

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Xzavier Shelton

Answered 2022-06-25
Author has **26** answers

${\mathrm{cos}}^{2}x>{\mathrm{sin}}^{2}x\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x>0$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}2n\pi -{\displaystyle \frac{\pi}{2}}<2x<2n\pi +{\displaystyle \frac{\pi}{2}}$

where n is any integer

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}2n\pi -{\displaystyle \frac{\pi}{2}}<2x<2n\pi +{\displaystyle \frac{\pi}{2}}$

where n is any integer

Hector Petersen

Answered 2022-06-26
Author has **6** answers

my approach is to divide both sides by ${\mathrm{cos}}^{2}(x)$ to get

${\mathrm{tan}}^{2}(x)<1$

or equivalently

$-1<\mathrm{tan}(x)<1$

Remember that $\mathrm{tan}(x+\pi )=\mathrm{tan}(x)$

${\mathrm{tan}}^{2}(x)<1$

or equivalently

$-1<\mathrm{tan}(x)<1$

Remember that $\mathrm{tan}(x+\pi )=\mathrm{tan}(x)$

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The counterpart questions for sine and tangent can be handled as follows:

If$\frac{\mathrm{sin}A}{2}=\frac{\mathrm{sin}B}{3}=\frac{\mathrm{sin}C}{7}$ we can rule out triangle because by the Sine Rule a=2k, b=3k, c=7k $\Rightarrow a+b<c$

If$\frac{\mathrm{tan}A}{2}=\frac{\mathrm{tan}B}{3}=\frac{\mathrm{tan}C}{7}$ , we can see that a triangle will be made as $\mathrm{tan}A=2k,\mathrm{tan}B=3k,\mathrm{tan}C=7k$ , when inserted in the identity $\mathrm{tan}A+\mathrm{tan}B+\mathrm{tan}C=\mathrm{tan}A\mathrm{tan}B\mathrm{tan}C\Rightarrow k=\sqrt{\frac{2}{7}}$

If

If

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