• sin x/2 =
• cos x/2 =
• tan x/2 =
sin(x) = 7/17
cos(x) = sqrt(1 - sin(x)^2) = sqrt(1 - 49/289) = sqrt(240 / 289) = sqrt(16 * 15 / 289) = (4/17) * sqrt(15)
sin(x/2) =>
sqrt((1/2) * (1 - cos(x))) =>
sqrt((1/2) * (1 - (4/17) * sqrt(15))) =>
sqrt((2/4) * (17 - 4 * sqrt(15)) / 17) =>
sqrt((34 / (4 * 17^2)) * (17 - 4 * sqrt(15))) =>
sqrt(34 * (17 - 4 * sqrt(15))) / 34
cos(x/2) =>
sqrt((1/2) * (1 + cos(x))) =>
sqrt((2/4) * (1 + (4/17) * sqrt(15))) =>
(1/17) * (1/2) * sqrt(2 * 17 * (17 + 4 * sqrt(15))) =>
sqrt(34 * (17 + 4 * sqrt(15))) / 34
tan(x/2) =>
sin(x/2) / cos(x/2) =>
sqrt(34 * (17 - 4 * sqrt(15))) / sqrt(34 * (17 + 4 * sqrt(15))) =>
sqrt((17 - 4 * sqrt(15)) / (17 + 4 * sqrt(15))) =>
(17 - 4 * sqrt(15)) / sqrt(289 - 16 * 15)) =>
(17 - 4 * sqrt(15)) / sqrt(49) =>
(17 - 4 * sqrt(15)) / 7
sin(x) = 7/17 â where 0° < x < 90° â cos(x) > 0
cos²(x) + sin²(x) = 1
cos²(x) = 1 - sin²(x)
cos²(x) = 1 - (7/17)²
cos²(x) = 1 - (49/289)
cos²(x) = (289/289) - (49/289)
cos²(x) = (289 - 49)/289
cos²(x) = 240/289 â recall cos(x) > 0
cos(x) = (â240)/(â289)
cos(x) = (â240)/17
cos(x) = [â(16 * 15)]/17
cos(x) = (4â15)/17
tan(x) = sin(x)/cos(x)
tan(x) = (7/17) / [(4â15)/17]
tan(x) = 7/(4â15)
Do you know this identity:
cos(2x) = 2cos²(x) - 1
Adapt this result to your case:
cos(x) = 2cos²(x/2) - 1
2cos²(x/2) = cos(x) - 1 â recall: cos(x) = (4â15)/17
2cos²(x/2) = [(4â15)/17] - 1
2cos²(x/2) = [(4â15)/17] - (17/17)
2cos²(x/2) = (4â15 - 17)/17
cos²(x/2) = (4â15 - 17)/34
cos(x/2) = â[(4â15 - 17)/34]
sin(2x) = 2sin(x).cos(x)
sin(x) = 2sin(x/2).cos(x/2)
2sin(x/2) = sin(x)/cos(x/2) â recall: sin(x) = 7/17
2sin(x/2) = (7/17)/cos(x/2) â recall: cos(x/2) = â[(4â15 - 17)/34]
2sin(x/2) = (7/17)/â[(4â15 - 17)/34]
2sin(x/2) = (7/17) * 1/â[(4â15 - 17)/34]
sin(x/2) = (7/34) * 1/â[(4â15 - 17)/34]
tan(2x) = 2tan(x) / [1 - tan²(x)]
tan(x) = 2tan(x/2) / [1 - tan²(x/2)]
Faster with:
tan(x/2) = sin(x/2) / cos(x/2)
tan(x/2) = { (7/34) * 1/â[(4â15 - 17)/34] } / { â[(4â15 - 17)/34] }
tan(x/2) = { (7/34) * 1/â[(4â15 - 17)/34] } * { 1/â[(4â15 - 17)/34] }
tan(x/2) = (7/34) * 1/[(4â15 - 17)/34]
tan(x/2) = 7/[(4â15 - 17)/34²]
tan(x/2) = (7 * 34²)/(4â15 - 17)
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Answers & Comments
Verified answer
sin(x) = 7/17
cos(x) = sqrt(1 - sin(x)^2) = sqrt(1 - 49/289) = sqrt(240 / 289) = sqrt(16 * 15 / 289) = (4/17) * sqrt(15)
sin(x/2) =>
sqrt((1/2) * (1 - cos(x))) =>
sqrt((1/2) * (1 - (4/17) * sqrt(15))) =>
sqrt((2/4) * (17 - 4 * sqrt(15)) / 17) =>
sqrt((34 / (4 * 17^2)) * (17 - 4 * sqrt(15))) =>
sqrt(34 * (17 - 4 * sqrt(15))) / 34
cos(x/2) =>
sqrt((1/2) * (1 + cos(x))) =>
sqrt((2/4) * (1 + (4/17) * sqrt(15))) =>
(1/17) * (1/2) * sqrt(2 * 17 * (17 + 4 * sqrt(15))) =>
sqrt(34 * (17 + 4 * sqrt(15))) / 34
tan(x/2) =>
sin(x/2) / cos(x/2) =>
sqrt(34 * (17 - 4 * sqrt(15))) / sqrt(34 * (17 + 4 * sqrt(15))) =>
sqrt((17 - 4 * sqrt(15)) / (17 + 4 * sqrt(15))) =>
(17 - 4 * sqrt(15)) / sqrt(289 - 16 * 15)) =>
(17 - 4 * sqrt(15)) / sqrt(49) =>
(17 - 4 * sqrt(15)) / 7
sin(x) = 7/17 â where 0° < x < 90° â cos(x) > 0
cos²(x) + sin²(x) = 1
cos²(x) = 1 - sin²(x)
cos²(x) = 1 - (7/17)²
cos²(x) = 1 - (49/289)
cos²(x) = (289/289) - (49/289)
cos²(x) = (289 - 49)/289
cos²(x) = 240/289 â recall cos(x) > 0
cos(x) = (â240)/(â289)
cos(x) = (â240)/17
cos(x) = [â(16 * 15)]/17
cos(x) = (4â15)/17
tan(x) = sin(x)/cos(x)
tan(x) = (7/17) / [(4â15)/17]
tan(x) = 7/(4â15)
Do you know this identity:
cos(2x) = 2cos²(x) - 1
Adapt this result to your case:
cos(x) = 2cos²(x/2) - 1
2cos²(x/2) = cos(x) - 1 â recall: cos(x) = (4â15)/17
2cos²(x/2) = [(4â15)/17] - 1
2cos²(x/2) = [(4â15)/17] - (17/17)
2cos²(x/2) = (4â15 - 17)/17
cos²(x/2) = (4â15 - 17)/34
cos(x/2) = â[(4â15 - 17)/34]
Do you know this identity:
sin(2x) = 2sin(x).cos(x)
Adapt this result to your case:
sin(x) = 2sin(x/2).cos(x/2)
2sin(x/2) = sin(x)/cos(x/2) â recall: sin(x) = 7/17
2sin(x/2) = (7/17)/cos(x/2) â recall: cos(x/2) = â[(4â15 - 17)/34]
2sin(x/2) = (7/17)/â[(4â15 - 17)/34]
2sin(x/2) = (7/17) * 1/â[(4â15 - 17)/34]
sin(x/2) = (7/34) * 1/â[(4â15 - 17)/34]
Do you know this identity:
tan(2x) = 2tan(x) / [1 - tan²(x)]
Adapt this result to your case:
tan(x) = 2tan(x/2) / [1 - tan²(x/2)]
Faster with:
tan(x/2) = sin(x/2) / cos(x/2)
tan(x/2) = { (7/34) * 1/â[(4â15 - 17)/34] } / { â[(4â15 - 17)/34] }
tan(x/2) = { (7/34) * 1/â[(4â15 - 17)/34] } * { 1/â[(4â15 - 17)/34] }
tan(x/2) = (7/34) * 1/[(4â15 - 17)/34]
tan(x/2) = 7/[(4â15 - 17)/34²]
tan(x/2) = (7 * 34²)/(4â15 - 17)