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Question-191144

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Question Number 191144 by mathlove last updated on 19/Apr/23
Answered by witcher3 last updated on 20/Apr/23
tan(a)tan(b)=((cos(a−b)−cos(a+b))/(cos(a−b)+cos(a+b))). tan(42)tan(78)=((cos(36)+cos(120))/(cos(36)−cos(120)))=((2cos(36)−1)/(2cos(36)+1)) tan(6)tan(66)=((cos(60)−cos(72))/(cos(60)+cos(72))) =((1−2cos(72))/(1+2cos(72))) ⇔((1−2cos(72))/(1+2cos(72))).((2cos(36)−1)/(2cos(36)+1))=1 cos(36)=a ⇔((1−2(2a^2 −1))/(1+2(2a^2 −1))).((2a−1)/(2a+1))=1 =((3−4a^2 )/((2a+1)^2 ))=1...(E) cos(2.36)=cos(3.36) ⇔2a^2 −1=4a^3 −3a ⇔4a^3 −2a^2 −3a+1=0 ⇔(4a−4)(a^2 +(a/2)−(1/4))=0 ⇒4a^2 +2a−1=0 3−4a^2 =2a+2 (2a+1)^2 =1+4a+4a^2 =1+4a+1−2a=2+2a (E)⇔((2a+2)/(2a+2))=1...True
$$\mathrm{tan}\left(\mathrm{a}\right)\mathrm{tan}\left(\mathrm{b}\right)=\frac{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)−\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)}. \\ $$$$\mathrm{tan}\left(\mathrm{42}\right)\mathrm{tan}\left(\mathrm{78}\right)=\frac{\mathrm{cos}\left(\mathrm{36}\right)+\mathrm{cos}\left(\mathrm{120}\right)}{\mathrm{cos}\left(\mathrm{36}\right)−\mathrm{cos}\left(\mathrm{120}\right)}=\frac{\mathrm{2cos}\left(\mathrm{36}\right)−\mathrm{1}}{\mathrm{2cos}\left(\mathrm{36}\right)+\mathrm{1}} \\ $$$$\mathrm{tan}\left(\mathrm{6}\right)\mathrm{tan}\left(\mathrm{66}\right)=\frac{\mathrm{cos}\left(\mathrm{60}\right)−\mathrm{cos}\left(\mathrm{72}\right)}{\mathrm{cos}\left(\mathrm{60}\right)+\mathrm{cos}\left(\mathrm{72}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{72}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{72}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{72}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{72}\right)}.\frac{\mathrm{2cos}\left(\mathrm{36}\right)−\mathrm{1}}{\mathrm{2cos}\left(\mathrm{36}\right)+\mathrm{1}}=\mathrm{1} \\ $$$$\mathrm{cos}\left(\mathrm{36}\right)=\mathrm{a} \\ $$$$\Leftrightarrow\frac{\mathrm{1}−\mathrm{2}\left(\mathrm{2a}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{1}+\mathrm{2}\left(\mathrm{2a}^{\mathrm{2}} −\mathrm{1}\right)}.\frac{\mathrm{2a}−\mathrm{1}}{\mathrm{2a}+\mathrm{1}}=\mathrm{1} \\ $$$$=\frac{\mathrm{3}−\mathrm{4a}^{\mathrm{2}} }{\left(\mathrm{2a}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}…\left(\mathrm{E}\right) \\ $$$$\mathrm{cos}\left(\mathrm{2}.\mathrm{36}\right)=\mathrm{cos}\left(\mathrm{3}.\mathrm{36}\right) \\ $$$$\Leftrightarrow\mathrm{2a}^{\mathrm{2}} −\mathrm{1}=\mathrm{4a}^{\mathrm{3}} −\mathrm{3a} \\ $$$$\Leftrightarrow\mathrm{4a}^{\mathrm{3}} −\mathrm{2a}^{\mathrm{2}} −\mathrm{3a}+\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{4a}−\mathrm{4}\right)\left(\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4a}^{\mathrm{2}} +\mathrm{2a}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}−\mathrm{4a}^{\mathrm{2}} =\mathrm{2a}+\mathrm{2} \\ $$$$\left(\mathrm{2a}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{4a}+\mathrm{4a}^{\mathrm{2}} =\mathrm{1}+\mathrm{4a}+\mathrm{1}−\mathrm{2a}=\mathrm{2}+\mathrm{2a} \\ $$$$\left(\mathrm{E}\right)\Leftrightarrow\frac{\mathrm{2a}+\mathrm{2}}{\mathrm{2a}+\mathrm{2}}=\mathrm{1}…\mathrm{True} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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