Question

Use the given information to find

​(a) sine (s plus t )sin(s+t)​, ​
(b) tangent (s plus t )tan(s+t)​, and
​(c) the quadrant of splus+t. sine s equals one seventhsins= 1 7 and sine t equals negative four seventhssint=− 4 7​, s in quadrant II and t in quadrant IV

Answer 1

(A). 0.6828

(B). 0.9346

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

Explanation:

hello,

i will use

`\sin \ s= (1)/(7)\\\sin \ t =(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 = ±
`√(1-\sin s)`

cos s = ±
`\sqrt{1- ((1)/(7))^(2) }`

cos s = ±
`\sqrt{(49)/(49) -(1)/(49) }`

cos s = ±
`\sqrt{(48)/(49) }`

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

cos s = -
`\sqrt{(48)/(49) }`

next we find cos t using the same method

cos t = ±
`√(1- \sin ^2 t)`

cos t =±
`\sqrt{1- ((-4)/(7))^2 }`

cos t = ±
`\sqrt{(49)/(49) - (16)/(49) }`

cos t = ±
`\sqrt{(33)/(49) }`

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

cos t =
`\sqrt{(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) =
`(1)/(7) \sqrt{(33)/(49) } \ + (-(4)/(7) ) (-\sqrt{(48)/(49) } )`

sin(S + t) =
`(√(33) )/(49) \ + (4√(48) )/(49)`

sin(S + t) =
`(√(33) \ + 4√(48) )/(49)`

= 0.6828

(B) please recall the trigonometric identity

`\tan (A+B) = (tan A\ + \ tan B )/(1- tan Atan B)` (1)

`\tan x = (\sin x)/(\cos x)`

thus

`\tan s = ((1)/(7) )/(-(√(48) )/(7) ) = -(1)/(√(48) )`

`\tan t = \frac{-(4)/(7) }{{(√(33) )/(7) } } =-(4)/(√(33) )`

applying (1) above we have

`\tan (s + t) = (-(1)/(√(48) ) -(4)/(√(33) ) )/(1- (-(1)/(√(48) )) (-(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.

Answer 2

This will guide you through

Explanation: