jika sudut A dikuadran ll dan b di kuadran lll . maka sin A=3\4 dan cosvB=-7/8.Tentukan sin(A+B) ,Cos(A-B), Tan(A+B)

[tex]cos \: a = \sqrt{1 - {sin}^{2} a} = \sqrt{1 - \frac{9}{16} } = - \frac{ \sqrt{7} }{4} (kuadran \: 2) \\ sin \: b = \sqrt{1 - {cos}^{2} b} = \sqrt{1 - \frac{49}{64} } = - \frac{ \sqrt{15} }{8} (kuadran \: 3)[/tex]
sin (a+b) = sin a cos b + cos a sin b =
[tex] \frac{3}{4} \times ( - \frac{7}{8} ) + \frac{ \sqrt{7} }{4} \times \frac{ \sqrt{15} }{8} = \frac{ - 21 + \sqrt{105} }{32} [/tex]
cos (a - b) = cos a cos b + sin a sin b =
[tex] \frac{7 \sqrt{7} }{32} - \frac{ \sqrt{15} }{32} = \frac{7 \sqrt{7} - \sqrt{15} }{32} [/tex]
tan (a + b) =
[tex] \frac{tan \: a \: + tan \: b}{1 - tan \: a \: tan \: b} = \frac{ - \frac{3}{ \sqrt{7}} + \frac{ \sqrt{15} }{7} }{1 + \frac{3 \sqrt{15} }{7 \sqrt{7} } } = \frac{ \frac{ \sqrt{105} - 21}{7 \sqrt{7} } }{ \frac{7 \sqrt{7} + 3 \sqrt{15} }{7 \sqrt{7} } } = \frac{ \sqrt{105} - 21}{7 \sqrt{7} + 3 \sqrt{15} } [/tex]