find an exact value for sin (3pi/8)

2 answers

Remark :
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pi / 8 = 22.5° = 22° 30'

3 pi / 8 = 4 pi / 8 - pi / 8 = pi / 2 - pi / 8

3 pi / 8 = 90° - 22° 30' = 67° 30'

pi / 4 = 45°

sin ( pi / 4 ) = sqroot ( 2 ) / 2

cos ( pi / 4 ) = sqroot ( 2 ) / 2

pi / 2 = 90°

sin ( pi / 2 ) = 1

cos ( pi / 2 )= 0
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3 pi / 8 = pi / 2 - pi / 8

sin ( A - B ) = sin ( A ) cos ( B ) - cos ( A ) sin ( B )

sin ( 3 pi / 8 ) =

sin ( pi / 2 - pi / 8 ) =

sin ( pi / 2 ) cos ( pi / 8 ) - cos ( pi / 2 ) sin ( pi / 8 ) =

1 * cos ( pi / 8 ) - 0 * sin ( pi / 8 ) =

cos ( pi / 8 )

So :

sin ( 3 pi / 8 ) = cos ( pi / 8 )

pi / 8 = ( pi / 4 ) / 2

cos ( A / 2 ) = + OR - sqroot [ ( 1 + cos A ) / 2 ]

cos ( pi / 8 ) = cos ( pi / 4 ) / 2 = + OR - sqroot [ ( 1 + cos ( pi / 4 ) ) / 2 ] =

+ OR - sqroot [ ( 1 + sqroot ( 2 ) / 2 ) / 2 ] =

+ OR - sqroot [ ( 2 / 2 + sqroot ( 2 ) / 2 ) / 2 ] =

+ OR - sqroot [ ( ( 2 + sqroot ( 2 ) ) / 2 ) / 2 ] =

+ OR - sqroot [ ( 2 + sqroot ( 2 ) ) / ( 2 * 2 ) ] =

+ OR - sqroot [ ( 2 + sqroot ( 2 ) ) / 4 ] =

+ OR - sqroot [ 2 + sqroot ( 2 ) ] / sqroot ( 4 ) =

+ OR - sqroot [ 2 + sqroot ( 2 ) ] / 2

3pi / 8 is in quadrant I

In quadrant I cosine in positive so :

cos ( pi / 8 ) = + sqroot [ 2 + sqroot ( 2 ) ] / 2 = sqroot [ 2 + sqroot ( 2 ) ] / 2

We alredy know :

sin ( 3 pi / 8 ) = cos ( pi / 8 ) so :

sin ( 3 pi / 8 ) = sqroot [ 2 + sqroot ( 2 ) ] / 2
P.S.

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sin ( 3 pi / 8 )

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