a) prove that sin^2 ∮ = 1 over 2
更新1:
It is given that 0<&<360, and x^2+3x sin& - cos^2 & =0 is an equation in x such that the sum of the squares of the two roots is 11 over 2 . a) prove that sin^2 & = 1 over 2
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2010-04-03 1:13 am
It is given that 0<∮<360, and x^2+3x sin∮ - cos^2 ∮ =0 is an equation in x such that the sum of the squares of the two roots is 11 over 2 .
a) prove that sin^2 ∮ = 1 over 2
a) prove that sin^2 ∮ = 1 over 2
回答 (1)
2010-04-03 1:22 am
✔ 最佳答案
x^2 + 3x sin@ - cos^2 @ = 0 , ( 0 < @ < 360 )the sum of the squares of the two roots
= a^2 + b^2 = (a^2 + 2ab + b^2 - 2ab) = (a + b)^2 - 2ab = 11/2......*
Sum of the roots = a + b = - 3sin@
Product of the roots = ab = - cos^2 @
Sub to * :
(- 3sin@)^2 - 2( - cos^2 @) = 11/2
9(sin@)^2 + 2(cos@)^2 = 11/2
7(sin@)^2 + 2(sin@)^2 + 2(cos@)^2 = 11/2
7(sin@)^2 + 2[(sin@)^2 + 2(cos@)^2] = 11/2
7(sin@)^2 + 2(1) = 11/2
7(sin@)^2 = 11/2 - 4/2 = 7/2
(sin@)^2 = 1/2
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