It is known that sin Θ and cos Θ are roots of the equation x ^ 2-2xsina + sin ^ 2B = 0 derived from X

It is known that sin Θ and cos Θ are roots of the equation x ^ 2-2xsina + sin ^ 2B = 0 derived from X

sin^2Θ+cos^2Θ=（sinΘ+cosΘ) ^2-2sinΘcosΘ=1
Translation according to Weida's theorem
sinΘ+cosΘ=-2sina
sinΘcosΘ=sin^2b
So (sin Θ + cos Θ) ^ 2-2sin Θ cos Θ = 4sin ^ 2a-2sin ^ B = 1
4sin^2a-2sin^2b=1
-1-2sin^2b=-4sin^a
1-2sin^2b=2-4sin^a=2(1-2sin^a)
cos2b=2cos2a
Get proof

The proof of (1-cos square α) / (sin α - cos α) - (sin α + cos α) / Tan square α - 1 = sin α + cos α

It is proved that: left = (1-cos & # 178; α) / (sin α - cos α) - (sin α + cos α) / [(Sin & # 178; α - cos & # 178; α) / cos & # 178; α] = Sin & # 178; α / (sin α - cos α) - cos & # 178; α (sin α + cos α) / (Sin & # 178; α - cos & # 178; α) = Sin & # 178; α / (sin α - cos α) / (sin α - cos α) / (sin α - cos α) / (sin α - cos α) / (Sin & # 178; α) = (Sin & # 178; α / (sin α - cos α) / (sin α - cos α) / (sin α - cos α) / (sin α) / (sin α - cos α) / (sin α) / (sin α - cos α) / (

Verification: 3 / sin squared 40 degrees minus 1 / cos squared 40 degrees = 32sin10 degrees

The original formula = (√ 3cos40-sin40)) (√ 3cos40 + 40) / sin square40cos square40
=16 (√ 3cos40 / 2-sin40 / 2) (√ cos40 / 2 + sin40 / 2) / sin80
=16sin20cos10 / cos square 10
=16*2sin10cos10/cos10
=32sin10

Given sin (a + π) = 5 / 4 and sinacosa < 0, find 2Sin (a - π) + 3tan (3 π - a) / 4cos (A-3 π)

Sin (a + π) = 4 / 5-sina = 4 / 5sina = - 4 / 5sinacosa < 0, then cosa > 0sin & # 178; a + cos & # 178; a = 1: cosa = 3 / 5tana = - 4 / 32sin (a - π) + 3tan (3 π - a) / 4cos (A-3 π) = - 2sina + 3tana / 4cosa = 8 / 5 + (- 4) / (12 / 15) = 8 / 5-5 = - 17 / 5