Find $\mathrm{cos}8x$ if:

$\mathrm{tan}x\mathrm{tan}2x=\mathrm{cot}2x\mathrm{cot}3x$

$\mathrm{tan}x\mathrm{tan}2x=\mathrm{cot}2x\mathrm{cot}3x$

Extrakt04
2022-06-30
Answered

Find $\mathrm{cos}8x$ if:

$\mathrm{tan}x\mathrm{tan}2x=\mathrm{cot}2x\mathrm{cot}3x$

$\mathrm{tan}x\mathrm{tan}2x=\mathrm{cot}2x\mathrm{cot}3x$

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frethi38

Answered 2022-07-01
Author has **16** answers

Let $t:=\mathrm{tan}x$ and $T:={t}^{2}\ge 0$ (assuming $x\in \mathbb{R}$), so

$\frac{4{T}^{2}(3-T)}{(1-T{)}^{2}(1-3T)}=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(T+1)({T}^{2}-6T+1)=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{T}^{2}-6T+1=0,$

and

$\mathrm{cos}4x=2{\left(\frac{1-T}{1+T}\right)}^{2}-1=\frac{{T}^{2}-6T+1}{(1+T{)}^{2}}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}8x=-1.$

$\frac{4{T}^{2}(3-T)}{(1-T{)}^{2}(1-3T)}=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(T+1)({T}^{2}-6T+1)=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{T}^{2}-6T+1=0,$

and

$\mathrm{cos}4x=2{\left(\frac{1-T}{1+T}\right)}^{2}-1=\frac{{T}^{2}-6T+1}{(1+T{)}^{2}}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}8x=-1.$

Amber Quinn

Answered 2022-07-02
Author has **3** answers

Using the product-to-sum identity $\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}\theta \mathrm{tan}\phi =\frac{\mathrm{cos}(\theta -\phi )-\mathrm{cos}(\theta +\phi )}{\mathrm{cos}(\theta -\phi )+\mathrm{cos}(\theta +\phi )}\phantom{\rule{thinmathspace}{0ex}}$

$\begin{array}{}\text{(1)}& 1={\mathrm{tan}}^{2}2x\cdot \frac{\mathrm{cos}2x-\mathrm{cos}4x}{\mathrm{cos}2x+\mathrm{cos}4x}=\frac{1-{\mathrm{cos}}^{2}2x}{{\mathrm{cos}}^{2}2x}\cdot \frac{\mathrm{cos}2x-2{\mathrm{cos}}^{2}2x+1}{\mathrm{cos}2x+2{\mathrm{cos}}^{2}2x-1}\end{array}$

With $\phantom{\rule{thinmathspace}{0ex}}t=\mathrm{cos}2x\phantom{\rule{thinmathspace}{0ex}}$

$\begin{array}{rl}{t}^{2}(2{t}^{2}+t-1)=(1-{t}^{2})(-2{t}^{2}+t+1)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}2{t}^{3}+2{t}^{2}-t-1=0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(t+1)(2{t}^{2}-1)=0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}t\in \{-1,\pm \frac{\sqrt{2}}{2}\}\end{array}$

The root $\phantom{\rule{thinmathspace}{0ex}}t=-1\phantom{\rule{thinmathspace}{0ex}}$ must be excluded because the denominator in (1) vanishes when $\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}2x=-1\phantom{\rule{thinmathspace}{0ex}}$, which leaves $\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}2x=\pm \frac{\sqrt{2}}{2}\phantom{\rule{thinmathspace}{0ex}}$

$\begin{array}{}\text{(1)}& 1={\mathrm{tan}}^{2}2x\cdot \frac{\mathrm{cos}2x-\mathrm{cos}4x}{\mathrm{cos}2x+\mathrm{cos}4x}=\frac{1-{\mathrm{cos}}^{2}2x}{{\mathrm{cos}}^{2}2x}\cdot \frac{\mathrm{cos}2x-2{\mathrm{cos}}^{2}2x+1}{\mathrm{cos}2x+2{\mathrm{cos}}^{2}2x-1}\end{array}$

With $\phantom{\rule{thinmathspace}{0ex}}t=\mathrm{cos}2x\phantom{\rule{thinmathspace}{0ex}}$

$\begin{array}{rl}{t}^{2}(2{t}^{2}+t-1)=(1-{t}^{2})(-2{t}^{2}+t+1)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}2{t}^{3}+2{t}^{2}-t-1=0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(t+1)(2{t}^{2}-1)=0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}t\in \{-1,\pm \frac{\sqrt{2}}{2}\}\end{array}$

The root $\phantom{\rule{thinmathspace}{0ex}}t=-1\phantom{\rule{thinmathspace}{0ex}}$ must be excluded because the denominator in (1) vanishes when $\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}2x=-1\phantom{\rule{thinmathspace}{0ex}}$, which leaves $\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}2x=\pm \frac{\sqrt{2}}{2}\phantom{\rule{thinmathspace}{0ex}}$

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