Let theta be an angle in quadrant II such that $$\mathrm{cos}\theta =\frac{-8}{9}.$$

Julius Blankenship
2022-09-11
Answered

Let theta be an angle in quadrant II such that $$\mathrm{cos}\theta =\frac{-8}{9}.$$

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Solve the following equation

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Here is the problem that I am struggling with it I tried to take a logarithm of both side but I kinda stuck can someone help me with this? Thanks

Here is the problem that I am struggling with it I tried to take a logarithm of both side but I kinda stuck can someone help me with this? Thanks

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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 $\mathrm{tan}A+\mathrm{tan}B+\mathrm{tan}C=\mathrm{tan}A\mathrm{tan}B\mathrm{tan}C\Rightarrow k=\sqrt{\frac{2}{7}}$

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If

asked 2022-05-21

Show trigonometric identity $\frac{1-\mathrm{tan}\frac{\theta}{2}}{1+\mathrm{tan}\frac{\theta}{2}}=\mathrm{sec}\theta -\mathrm{tan}\theta $