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Find the value of y such that $$\tan(y°)=(4\cos^29°-3)(4\cos^227°-3)$$

My work

I let $x = 9°$ And equation become $$(2\cos2x-1)(2\cos6x-1)$$

Then I tried to convert cos in tan using half angle formula but it didn't work .

I also found $\cos(4x+6x)=0 \implies \tan4x\tan6x=1$ But I failed to convert in a form from where I can calculate y.

Thanks in advance for any help

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$$\cos3x=\cos x(4\cos^2x-3)$$

For $\cos x\ne0,$

$$4\cos^2x-3=\dfrac{\cos 3x}{\cos x}$$

$$(4\cos^29^\circ-3)(4\cos^227^\circ-3)=\dfrac{\cos27^\circ}{\cos9^\circ}\cdot\dfrac{\cos81^\circ}{\cos27^\circ}$$

Finally $\cos81^\circ=\sin?^\circ$

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  • $\begingroup$ Sorry sir for my ignorance ! But can you explain how $$cos4x=cosx(4cos^2x-3)$$ $\endgroup$ – Rishi Nov 11 '19 at 10:06
  • $\begingroup$ @Rishi, See proofwiki.org/wiki/Triple_Angle_Formulas/Cosine#Proof_1 $\endgroup$ – lab bhattacharjee Nov 11 '19 at 10:08
  • $\begingroup$ Thank you I know formula , but got confused when you wrote 4x , thanks $\endgroup$ – Rishi Nov 11 '19 at 10:11
  • $\begingroup$ @Rishi, Sorry for the horrible typo $\endgroup$ – lab bhattacharjee Nov 11 '19 at 10:20

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