[toan11]phương trình lượng giác

M

mika_tmc

Mình ghj lại đề cho dễ nhìn :D
Cho [TEX]sin^2a+sin^2b+sin^2c=1[/TEX] (1)
CM: [TEX]tan^2atan^2b+tan^2btan^2c+tan^2ctan^2a=1-2tan^2atan^2btan^2c[/TEX]

Cách này hơi dài :D Mình ghj ngắn gọn lại...
Ta có:
(1)\Leftrightarrow [TEX]sin^2a + sin^2b + sin^2c = 1 + sin^2sin^2b+sin^2bsin^2c+sin^2csin^2a-sin^2sin^2b-sin^2bsin^2c-sin^2csin^2a+sin^2a sin^2b sin^2c - sin^2a sin^2b sin^2c[/TEX]

\Leftrightarrow [TEX]sin^2sin^2b+sin^2bsin^2c+sin^2csin^2a = (1- sin^2c) - (sin^2a-sin^2asin^2c) - (sin^2b+sin^2bsin^2c) - (sin^2asin^2b+sin^2a sin^2b sin^2c)-sin^2a sin^2b sin^2c[/TEX]

\Leftrightarrow [TEX]sin^2sin^2b+sin^2bsin^2c+sin^2csin^2a = (1-sin^2a)(1-sin^2b)(1-sin^2c)+sin^2a sin^2b sin^2c[/TEX]

\Leftrightarrow [TEX]sin^2a sin^2b cos^2c + cos^2a sin^2b sin^2c + sin^2a cos^2b sin^2c +3sin^2a sin^2b sin^2c= cos^2a cos^2b cos^2c+sin^2a sin^2b sin^2c[/TEX]

\Leftrightarrow[TEX]tan^2atan^2b+tan^2btan^2c+tan^2ctan^2a=1-2tan^2atan^2btan^2c[/TEX]
 
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