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prove sin(3/2 x)*sin x / sin(1/2 x) = sin x + sin 2x

Guest Jul 28, 2017
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 #1
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Verify the following identity:
sin((3 x)/2) (sin(x))/(sin(x/2)) = sin(x) + sin(2 x)

Multiply both sides by sin(x/2):
sin(x) sin((3 x)/2) = ^?sin(x/2) (sin(x) + sin(2 x))

sin(x) sin((3 x)/2) = 1/2 (cos(x - (3 x)/2) - cos(x + (3 x)/2)) = 1/2 (cos(-x/2) - cos((5 x)/2)):
(cos(-x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) (sin(x) + sin(2 x))

Use the identity cos(-x/2) = cos(x/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) (sin(x) + sin(2 x))

sin(x/2) (sin(x) + sin(2 x)) = sin(x/2) sin(x) + sin(x/2) sin(2 x):
(cos(x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) sin(x) + sin(x/2) sin(2 x)

sin(x/2) sin(x) = 1/2 (cos(x/2 - x) - cos(x/2 + x)) = 1/2 (cos(-x/2) - cos((3 x)/2)):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(-x/2) - cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

Use the identity cos(-x/2) = cos(x/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2) - cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

(cos(x/2) - cos((3 x)/2))/(2) = 1/2 cos(x/2) - 1/2 cos((3 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

sin(x/2) sin(2 x) = 1/2 (cos(x/2 - 2 x) - cos(x/2 + 2 x)) = 1/2 (cos(-(3 x)/2) - cos((5 x)/2)):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos(-(3 x)/2) - cos((5 x)/2))/(2)

Use the identity cos(-(3 x)/2) = cos((3 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2) - cos((5 x)/2))/(2)

(cos((3 x)/2) - cos((5 x)/2))/(2) = 1/2 cos((3 x)/2) - 1/2 cos((5 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2))/(2) - (cos((5 x)/2))/(2)

(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2))/(2) - (cos((5 x)/2))/(2) = 1/2 cos(x/2) - 1/2 cos((5 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((5 x)/2))/(2)

1/2 (cos(x/2) - cos((5 x)/2)) = 1/2 cos(x/2) - 1/2 cos((5 x)/2):
1/2 cos(x/2) - 1/2 cos((5 x)/2) = ^?1/2 cos(x/2) - 1/2 cos((5 x)/2)

The left hand side and right hand side are identical:
Answer: | (identity has been verified)

Guest Jul 28, 2017
 #2
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Using the trig identity

\(\displaystyle \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\) ,

 

RHS = \(\displaystyle \sin 2x + \sin x = 2 \sin \left( \frac{3x}{2}\right)\cos\left(\frac{x}{2}\right)= \frac{2 \sin \left( \frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} =\frac{\sin \left( \frac{3x}{2}\right)\sin\left(x\right)}{\sin\left(\frac{x}{2}\right)} \) .

 

Tiggsy

Guest Jul 28, 2017

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