tolong ya jawabkan nomor 2 3 4 5

TriGonoMetRi

2][
cos (x/2) = 12/13
sin (x/2) = 5/13
tan (x/2) = 5/12
(Triple Pythagoras)

sin x = sin 2(x/2)
= 2 sin (x/2) cos (x/2)
= 2 . 5/13 . 12/13
= 120/169

cos x = cos 2(x/2) = cos² x - sin² x
= (12/13)² - (5/13)²
= 119/169

tan x = sin x/cos x = 120/169 × 169/119 = 120/119

• x terletak di kuadran I (sin x dan cos x , positif)

3][
cos t = 3/5
cos 2(t/2) = 2 cos² (t/2) - 1
cos² (t/2) = (1 + 3/5)/2 = 8/10
cos (t/2) = √(8/10)

sin t = 4/5
sin 2(t/2) = 2 sin (t/2) cos (t/2)
4/5 = 2 . sin (t/2) . √(8/10)
sin (t/2) = 2/5 √(10/8)

sin (3/2)t
= sin (t + t/2)
= sin t cos t/2 + cos t sin t/2
= 4/5 . √(8/10) + 3/5 . 2/5 √(10/8)
= 4/5 . √(8/10) + 6/25 . √(10/8)
= 16/50 √5 + 3/25 √5
= 1/50 (16√5 + 6√5)
= 1/50 (22√5)


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2 aja ya.. :)