Answer:
(A). 0.6828
(B). 0.9346
(C). sin (s + t) lies in the first quadrant
Step-by-step explanation:
hello,
i will use
[tex]\sin \ s= \frac{1}{7}\\\sin \ t =\frac{4}{7}[/tex]
where S and t are in the third and fourth quadrant respectively.
next we find the value of cos s and cos t.
please recall that
cos x = ±[tex]\sqrt{1-\sin^{2} x }[/tex]
thus we have ;
cos s = ±[tex]\sqrt{1-\sin s}[/tex]
cos s = ±[tex]\sqrt{1- (\frac{1}{7})^{2} }[/tex]
cos s = ±[tex]\sqrt{\frac{49}{49} -\frac{1}{49} }[/tex]
cos s = ±[tex]\sqrt{\frac{48}{49} }[/tex]
since s is in the second quadrant, we choose the negative.
cos s = -[tex]\sqrt{\frac{48}{49} }[/tex]
next we find cos t using the same method
cos t = ±[tex]\sqrt{1- \sin ^2 t}[/tex]
cos t =±[tex]\sqrt{1- (\frac{-4}{7})^2 }[/tex]
cos t = ±[tex]\sqrt{\frac{49}{49} - \frac{16}{49} }[/tex]
cos t = ±[tex]\sqrt{\frac{33}{49} }[/tex]
since t is in the fourth quadrant, we choose the positive.
cos t = [tex]\sqrt{\frac{33}{49} }[/tex]
please recall the trigonometric identity
(A) sin(A+B) = sin A cos B + sin B cos A
sin(S + t) = sin S cos t + sin t cos S
sin(S + t) =[tex]\frac{1}{7} \sqrt{\frac{33}{49} } \ + (-\frac{4}{7} ) (-\sqrt{\frac{48}{49} } )[/tex]
sin(S + t) = [tex]\frac{\sqrt{33} }{49} \ + \frac{4\sqrt{48} }{49}[/tex]
sin(S + t) = [tex]\frac{\sqrt{33} \ + 4\sqrt{48} }{49}[/tex]
= 0.6828
(B) please recall the trigonometric identity
[tex]\tan (A+B) = \frac{tan A\ + \ tan B }{1- tan Atan B}[/tex] (1)
[tex]\tan x = \frac{\sin x}{\cos x}[/tex]
thus
[tex]\tan s = \frac{\frac{1}{7} }{-\frac{\sqrt{48} }{7} } = -\frac{1}{\sqrt{48} }[/tex]
[tex]\tan t = \frac{-\frac{4}{7} }{{\frac{\sqrt{33} }{7} } } =-\frac{4}{\sqrt{33} }[/tex]
applying (1) above we have
[tex]\tan (s + t) = \frac{-\frac{1}{\sqrt{48} } -\frac{4}{\sqrt{33} } }{1- (-\frac{1}{\sqrt{48} }) (-\frac{4}{33} )}[/tex]
= 0.9346
(c) sin (s + t) lies in the first quadrant because its value is a positive number and sine is positive in the first or second quadrant.
