Question

Use the given information to find

Answer 1

Answer:

This will guide you through

Step-by-step explanation:

Answer 2

Answer:

(A). 0.6828

(B). 0.9346

(C). sin (s + t) lies in the first quadrant

Step-by-step explanation:

hello,

i will use

$\sin \ s= \frac{1}{7}\\\sin \ t =\frac{4}{7}$

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 = ±$\sqrt{1-\sin^{2} x }$

thus we have ;

cos s = ±$\sqrt{1-\sin s}$

cos s  = ±$\sqrt{1- (\frac{1}{7})^{2} }$

cos s    = ±$\sqrt{\frac{49}{49} -\frac{1}{49} }$

cos s     = ±$\sqrt{\frac{48}{49} }$

since s is in the second quadrant, we choose the negative.

cos s = -$\sqrt{\frac{48}{49} }$

next we find cos t using the same method

cos t = ±$\sqrt{1- \sin ^2 t}$

cos t =±$\sqrt{1- (\frac{-4}{7})^2 }$

cos t = ±$\sqrt{\frac{49}{49} - \frac{16}{49} }$

cos t = ±$\sqrt{\frac{33}{49} }$

since t is in the fourth quadrant, we choose the positive.

cos t = $\sqrt{\frac{33}{49} }$

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)  =$\frac{1}{7} \sqrt{\frac{33}{49} } \ + (-\frac{4}{7} ) (-\sqrt{\frac{48}{49} } )$

    sin(S + t)   = $\frac{\sqrt{33} }{49} \ + \frac{4\sqrt{48} }{49}$

   sin(S + t)     = $\frac{\sqrt{33} \ + 4\sqrt{48} }{49}$

                      = 0.6828

(B) please recall the trigonometric identity

$\tan (A+B) = \frac{tan A\ + \ tan B }{1- tan Atan B}$        (1)

$\tan x = \frac{\sin x}{\cos x}$

thus

$\tan s = \frac{\frac{1}{7} }{-\frac{\sqrt{48} }{7} } = -\frac{1}{\sqrt{48} }$

$\tan t = \frac{-\frac{4}{7} }{{\frac{\sqrt{33} }{7} } } =-\frac{4}{\sqrt{33} }$

applying (1) above we have

$\tan (s + t) = \frac{-\frac{1}{\sqrt{48} } -\frac{4}{\sqrt{33} } }{1- (-\frac{1}{\sqrt{48} }) (-\frac{4}{33} )}$

                = 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.