Teigonometri
90° < a < 180° kuadran II
sin a = 5/13
cos a = - √(1 - sin² a)
cos a = - √(1 - (5/13)²)
cos a = - √((13²- 25)/13²)
cos a = - 1/13 √144
cos a = - 12/13
atau
tripel pythagoras → 5, 12, 13
tan (1/2 a)
= sin a / (1 + cos a)
= (5/13) / (1 - 12/13)
= (5/13) / (1/13)
= 5/1
= 5
= (1 - cos a)/sin a
= (1 - (-12/13))/(5/13)
= (25/13) / (5/13)
= 25/5
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Verified answer
Teigonometri
90° < a < 180° kuadran II
sin a = 5/13
cos a = - √(1 - sin² a)
cos a = - √(1 - (5/13)²)
cos a = - √((13²- 25)/13²)
cos a = - 1/13 √144
cos a = - 12/13
atau
tripel pythagoras → 5, 12, 13
tan (1/2 a)
= sin a / (1 + cos a)
= (5/13) / (1 - 12/13)
= (5/13) / (1/13)
= 5/1
= 5
atau
tan (1/2 a)
= (1 - cos a)/sin a
= (1 - (-12/13))/(5/13)
= (25/13) / (5/13)
= 25/5
= 5