Use a double-angle formula to rewrite the expression. 14 sin^2 x − 7?
回答 (7)
2016-07-31 5:25 pm
Trigonometric identity : cos(2x) = 1 - 2 sin²(x)
14 sin²(x) - 7
= - 7 [1 - 2 sin²(x)]
= - 7 cos(2x)
14 sin²(x) - 7
= - 7 [1 - 2 sin²(x)]
= - 7 cos(2x)
2016-07-31 5:21 pm
By double-angle identity, sin²x = {1 - cos(2x)}/2
So, 14*sin²x - 7 = 14*{1 - cos(2x)}/2 - 7 = 7 - 7*cos(2x) - 7 = -7*cos(2x)
So, 14*sin²x - 7 = 14*{1 - cos(2x)}/2 - 7 = 7 - 7*cos(2x) - 7 = -7*cos(2x)
2016-08-16 4:30 pm
14sin²x -7 = 7(2sin²x-1)
= -7(1-2sin²x) = -7cos(2x)
= -7(1-2sin²x) = -7cos(2x)
2016-07-31 8:01 pm
14sin^2 x -7 = 14(1-2sin^2x)-7
14 -28sin^2x - 7
-28sin^2 x - 7
14 -28sin^2x - 7
-28sin^2 x - 7
2016-07-31 6:38 pm
7 [ 1 - cos 2x ] - 7 = 7 [ 1 - cos 2x - 1 ]
7 [ 1 - cos 2x ] = - 7 cos 2x
7 [ 1 - cos 2x ] = - 7 cos 2x
2016-07-31 5:43 pm
-7cos(2x)
2016-07-31 5:25 pm
cos2x = 1 - 2sin²x
so, 2sin²x - 1 = -cos2x
=> 14sin²x - 7 = -7cos2x
:)>
so, 2sin²x - 1 = -cos2x
=> 14sin²x - 7 = -7cos2x
:)>
2016-07-31 5:23 pm
14sin²x -7 = 7(2sin²x-1)
= -7(1-2sin²x) = -7cos(2x)
= -7(1-2sin²x) = -7cos(2x)
收錄日期: 2021-05-01 13:05:43
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https://hk.answers.yahoo.com/question/index?qid=20160731091614AAZUA9g