Pre-Cal

Find sin x/2, cos x/2, and tan x/2 from the given information.

[1]. sin x = 5/13, 0° < x < 90°

1). sin x/2 =________.

2). cos x/2 =________.

3). tan x/2 =________.

[Note: Plz help with these 3]

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  1. Use Trigonometric Identities :

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

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

    0° < x < 90° meen x lies in Quadrant I

    In Quadrant I sin x and cos x are positive so :

    sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]

    cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]

    If sin x = 5 / 13 then :

    cos x = sqroot [ 1 - ( sin x ) ^ 2 ]

    cos x = sqroot [ 1 - ( 5 / 13 ) ^ 2 ] =

    sqroot [ 1 - 25 / 169 ] =

    sqroot [ 169 / 169 - 25 / 169 ] =

    sqroot [ 144 / 169 ] = 12 / 13

    cos x = 12 / 13

    Now :

    sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]

    sin ( x / 2 ) = sqroot [ ( 1 - 12 / 13 ) / 2 ] =

    sqroot [ ( 13 / 13 - 12 / 13 ) / 2 ] =

    sqroot [ ( 1 / 13 ) / 2 ] =

    sqroot [ 1 / ( 2 * 13 ) ] =

    sqroot ( 1 / 26 ) = 1 / sqroot ( 26 )

    sin ( x / 2 ) = 1 / sqroot ( 26 )

    cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]

    cos ( x / 2 ) = sqroot [ ( 1 + 12 / 13 ) / 2 ] =

    sqroot [ ( 13 / 13 + 12 / 13 ) / 2 ] =

    sqroot [ ( 25 / 13 ) / 2 ] =

    sqroot [ 25 / ( 2 * 13 ) ] =

    sqroot ( 25 / 26 ) = 5 / sqroot ( 26 )

    cos ( x / 2 ) = 5 / sqroot ( 26 )

    tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )

    tan ( x / 2 ) = [ 1 / sqroot ( 26 ) ] / [ 5 / sqroot ( 26 ) ] =

    [ 1 * sqroot ( 26 ) ] / [ 5 * sqroot ( 26 ) ] = 1 / 5

    tan ( x / 2 ) = 1 / 5

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