given
1.) sin2x sin 3x. the answer i get is =1/2(cos5x-cosx) is that correct?
2.) sin^2x = A cos(2x) + B. the answer i get is =-1/2 (cos2x) + 1/2, correct?
need help here.
euler formula/calculus?
2015-12-12 4:00 pm
回答 (2)
2015-12-12 4:24 pm
✔ 最佳答案
1)sin a sin b = (1/2) ( cos (a-b) - cos(a+b) )
a = 2x
b= 3x
sin 2x sin 3x = (1/2) (cos ( -x) - cos(5x) )
= (1/2) ( cos (x) - cos( 5x) )
However, this is not the Euler's formula
2)
sin^2 x = (1- cos (2x) ) / 2
sin^2 x = 1/2 - (1/2) cos 2x
sin^2 x = (-1/2) cos 2x + 1/2
A = -1/2
B = 1/2
However, this is not the Euler's formula
Euler's formula is e^ix = cos (x) + i sin(x)
2015-12-12 4:44 pm
sinA⋅sinB = (-1/2)[cos(A+B) - cos(A-B)]
sin(2x)⋅sin(3x)
= (-1/2)[cos(5x) - cos(-x)]
= (-1/2)[cos(5x) - cos(x)]
>> (1/2)(cos5x-cosx)......something wrong.
sin²x = A cos(2x) + B
sin²x = [1 - cos(2x)]/2 = (-1/2)cos(2x) + (1/2)
>> -1/2(cos2x) + 1/2...........correct.
sin(2x)⋅sin(3x)
= (-1/2)[cos(5x) - cos(-x)]
= (-1/2)[cos(5x) - cos(x)]
>> (1/2)(cos5x-cosx)......something wrong.
sin²x = A cos(2x) + B
sin²x = [1 - cos(2x)]/2 = (-1/2)cos(2x) + (1/2)
>> -1/2(cos2x) + 1/2...........correct.
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