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July 2022 1 20 Report

Доказать тождество
[tex] \frac{tg3 \beta }{tg2 \beta } -1- \frac{tg \beta }{tg2 \beta } =tg \beta tg3 \beta [/tex]

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Verified answer

$\beta =b\\ 
 \frac{tg3b}{tg2b}-1-\frac{tgb}{tg2b}=tgb*tg3b\\\\
 tg3b=\frac{sin3b}{cos3b}=\frac{(4cos^2b-1)sinb}{4cos^3b-3cosb}\\\\
 tg2b=\frac{sin2b}{cos2b}=\frac{2sinb*cosb}{2cos^2b-1}\\\\
 \frac{tg3b}{tg2b} = \frac{8cos^4b-6cos^2b+1}{8cos^4b-6cos^2b}\\\\
 \frac{tgb}{tg2b}=\frac{2cos^2b-1}{2cos^2b}\\\\
 tgb*tg3b=\frac{ 3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\\\
 \frac{8cos^4b-6cos^2b+1}{8cos^4b-6cos^2b}-\frac{2cos^2b-1}{2cos^2b}-1=\frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\\\

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то есть надо теперь доказать это выражение
$\frac{8cos^4b-6cos^2b+1}{8cos^4b-6cos^2b}-\frac{2cos^2b-1}{2cos^2b}-1=\frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\\\ 
 \frac{8cos^4b-6cos^2b+1}{2cos^2b(4cos^2b-3)}-\frac{(2cos^2b-1)(4cos^2b-3)}{2cos^2b(4cos^2b-3)}$
$\frac{8cos^4b-6cos^2b+1}{2cos^2b(4cos^2b-3)}-\frac{(2cos^2b-1)(4cos^2b-3)}{2cos^2b(4cos^2b-3)}-\\
\frac{2cos^2b(4cos^2b-3)}{2cos^2b(4cos^2b-3)}=\frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\\\
 \frac{8cos^4b-6cos^2b+1-(2cos^2b-1)(4cos^2b-3)-2cos^2b(4cos^2b-3)}{2cos^2b(4cos^2b-3)}=\\\\
 \frac{6sin^2b-8sin^4b}{8cos^4b-6cos^2b}=\frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}$
то есть верно

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