mohon bantuannya nomor 6 yaa kak

tanx = 12/5
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12²+5²=13²
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jadi sinx = 12/13
       cosx = 5/13

siny = 6/10
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10²-6²=8²
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jadi cosy = 8/10


a. cos(x+y)
= cosx . cosy  -  sinx . siny
= (5/13) (8/10)  -  (12/13) (6/10)
= (4/13) - (36/65)
= -(16/65)

b. sin(x+y)
= sinx . cosy  +  cosx . siny
= (12/13) (8/10)  +  (5/13) (6/10)
= (48/65) + (3/13)
= 63/65

tanx= 12/5 (depan/samping) sudut lancip =>kuadran 1, semua positif
miring=√depan²+samping²
         =√12²+5²
         =√144+25
         =√169
         =13
cos x= samping/miring = 5/13
sin x= depan/miring =12/13

sin y = 6/10 (depan/miring)
samping=√miring²-depan²
             =√10²-6²
             =√100-36
             =√64
             =8
cos y= samping/miring =8/10
tan y= depan/samping= 6/8

ditanya:
a.) cos(x+y)=cosx·cosy-sinx·siny
                  =(5/13)·(8/10)-(12/13)·(6/10)
                  =40/130 - 72/130
                  =-32/130
                 =-16/65
b.) sin(x+y)= sinx·cosy+cosx·siny
                 =(12/13)·(8/10)+(5/13)·(6/10)
                =96/130 + 30/130
                =126/130
               =63/65