Asked by Anonymous
                Use the appropriate sum/difference formula to find the exact value of cos(105°). 
do we choose one, then make 1 side equal cos of 105°? I don't understand the question.
            
        do we choose one, then make 1 side equal cos of 105°? I don't understand the question.
Answers
                    Answered by
            Bosnian
            
    105 ° = 210 ° / 2
210 ° = 180 ° + 30 °
cos ( A + B ) = cos ( A ) * cos ( B ) - sin ( A ) * sin ( B )
sin 30 ° = 1 / 2
sin 180 ° = 0
cos 30 ° = sqrt ( 3 ) / 2
cos 180 ° = - 1
cos 210 ° = cos ( 180 ° + 30 ° ) = cos 180 ° * cos 30 ° - sin 180 ° * sin 30 ° =
( - 1 ) * sqrt ( 3 ) / 2 - 0 * 1 / 2 =
- sqrt ( 3 ) / 2 - 0
cos 210 ° = - sqrt ( 3 ) / 2
cos ( theta / 2 ) = ± sqrt [ ( 1 / 2 ) * ( 1 + cos ( theta )) ]
cos 105 ° = cos ( 210 ° / 2 ) =
± sqrt [ ( 1 / 2 ) * ( 1 + ( - sqrt ( 3 ) / 2 ) ] =
± sqrt [ ( 1 / 2 ) * ( 2 / 2 + ( - sqrt ( 3 ) / 2 ) ] =
± sqrt [ ( 1 / 2 ) * ( ( 2 - sqrt ( 3 ) / 2 ) ) ] =
± sqrt [ ( ( 2 - sqrt ( 3 ) ) / 4 ] =
± ( 1 / 2 ) sqrt [ 2 - sqrt ( 3 ) ] =
± sqrt [ 2 - sqrt ( 3 ) ] / 2
The angle 105 ° is in the II quadrant
In quadrant II cosine is negative.
So
cos ( 105 ° ) = - sqrt [ 2 - sqrt ( 3 ) ] / 2
    
210 ° = 180 ° + 30 °
cos ( A + B ) = cos ( A ) * cos ( B ) - sin ( A ) * sin ( B )
sin 30 ° = 1 / 2
sin 180 ° = 0
cos 30 ° = sqrt ( 3 ) / 2
cos 180 ° = - 1
cos 210 ° = cos ( 180 ° + 30 ° ) = cos 180 ° * cos 30 ° - sin 180 ° * sin 30 ° =
( - 1 ) * sqrt ( 3 ) / 2 - 0 * 1 / 2 =
- sqrt ( 3 ) / 2 - 0
cos 210 ° = - sqrt ( 3 ) / 2
cos ( theta / 2 ) = ± sqrt [ ( 1 / 2 ) * ( 1 + cos ( theta )) ]
cos 105 ° = cos ( 210 ° / 2 ) =
± sqrt [ ( 1 / 2 ) * ( 1 + ( - sqrt ( 3 ) / 2 ) ] =
± sqrt [ ( 1 / 2 ) * ( 2 / 2 + ( - sqrt ( 3 ) / 2 ) ] =
± sqrt [ ( 1 / 2 ) * ( ( 2 - sqrt ( 3 ) / 2 ) ) ] =
± sqrt [ ( ( 2 - sqrt ( 3 ) ) / 4 ] =
± ( 1 / 2 ) sqrt [ 2 - sqrt ( 3 ) ] =
± sqrt [ 2 - sqrt ( 3 ) ] / 2
The angle 105 ° is in the II quadrant
In quadrant II cosine is negative.
So
cos ( 105 ° ) = - sqrt [ 2 - sqrt ( 3 ) ] / 2
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