To prove that Cos 2B /b^2 - Cos 2A/ a^2 = a^2 - b^2 / a^2 b^2
To prove that Cos 2B /b^2 - Cos 2A/ a^2 = a^2 - b^2 / a^2 b^2
Samaira Shah , 8 Years ago
Grade 11
Trigonometry> To prove that Cos 2B /b^2 - Cos 2A/ a^2 =...
1 Answers
Arun
Hello
First use the formula cos 2A = 1 - 2sin²A
(1 - 2sin²A)/a² - (1 - 2sin²B)/b² = 1/a² - 2sin²A/a² - 1/b² + 2sin²B/b²
= 1/a² - 1/b² + 2sin²B/b² - 2sin²A/a²
Now think of the law of sines in a triangle
sinA/a = sinB/b
sin²A/a² = sin²B/b²
Hence
2sin²B/b² - 2sin²A/a² = 0
So
Cos2A/a² - cos2B/b² = 1/a² - 1/b²
First use the formula cos 2A = 1 - 2sin²A
(1 - 2sin²A)/a² - (1 - 2sin²B)/b² = 1/a² - 2sin²A/a² - 1/b² + 2sin²B/b²
= 1/a² - 1/b² + 2sin²B/b² - 2sin²A/a²
Now think of the law of sines in a triangle
sinA/a = sinB/b
sin²A/a² = sin²B/b²
Hence
2sin²B/b² - 2sin²A/a² = 0
So
Cos2A/a² - cos2B/b² = 1/a² - 1/b²
Hence
Cos2B/a² - cos2A/b² = 1/b² - 1/a² = (a² - b²)/a² b²
Hence proved
Regards
Arun (askIITians forum expert)
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