# Find cos &#x2061;<!-- ⁡ --> 8 x if: tan &#x2061;<!-- ⁡ --> x tan &#x2061;<!--

Extrakt04 2022-06-30 Answered
Find $\mathrm{cos}8x$ if:
$\mathrm{tan}x\mathrm{tan}2x=\mathrm{cot}2x\mathrm{cot}3x$
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## Answers (2)

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
$\frac{4{T}^{2}\left(3-T\right)}{\left(1-T{\right)}^{2}\left(1-3T\right)}=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(T+1\right)\left({T}^{2}-6T+1\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{T}^{2}-6T+1=0,$
and
$\mathrm{cos}4x=2{\left(\frac{1-T}{1+T}\right)}^{2}-1=\frac{{T}^{2}-6T+1}{\left(1+T{\right)}^{2}}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}8x=-1.$
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Amber Quinn
Answered 2022-07-02 Author has 3 answers
Using the product-to-sum identity $\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}\theta \mathrm{tan}\phi =\frac{\mathrm{cos}\left(\theta -\phi \right)-\mathrm{cos}\left(\theta +\phi \right)}{\mathrm{cos}\left(\theta -\phi \right)+\mathrm{cos}\left(\theta +\phi \right)}\phantom{\rule{thinmathspace}{0ex}}$
$\begin{array}{}\text{(1)}& 1={\mathrm{tan}}^{2}2x\cdot \frac{\mathrm{cos}2x-\mathrm{cos}4x}{\mathrm{cos}2x+\mathrm{cos}4x}=\frac{1-{\mathrm{cos}}^{2}2x}{{\mathrm{cos}}^{2}2x}\cdot \frac{\mathrm{cos}2x-2{\mathrm{cos}}^{2}2x+1}{\mathrm{cos}2x+2{\mathrm{cos}}^{2}2x-1}\end{array}$
With $\phantom{\rule{thinmathspace}{0ex}}t=\mathrm{cos}2x\phantom{\rule{thinmathspace}{0ex}}$
$\begin{array}{rl}{t}^{2}\left(2{t}^{2}+t-1\right)=\left(1-{t}^{2}\right)\left(-2{t}^{2}+t+1\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}2{t}^{3}+2{t}^{2}-t-1=0\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(t+1\right)\left(2{t}^{2}-1\right)=0\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}t\in \left\{-1,±\frac{\sqrt{2}}{2}\right\}\end{array}$
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 $\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}2x=±\frac{\sqrt{2}}{2}\phantom{\rule{thinmathspace}{0ex}}$
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