If sin s = -12/13sins=1213 and sin t = 4/5sint=45, what is cos(s+t)cos(s+t) and cos (s-t)cos(st)?

ss is in quadrant IV and tt in quadrant II.

1 Answer
May 25, 2018

cos(s+t) = 33/65cos(s+t)=3365

cos(s-t) = -63/65cos(st)=6365

Explanation:

We will need the values of cos(s) and cos(t)cos(s)andcos(t), therefore, we shall use the identity:

cos(x) = +-sqrt(1-sin^2(x))" [1]"cos(x)=±1sin2(x) [1]

Substitute sin(s)= -12/13sin(s)=1213 into equation [1]:

cos(s) = +-sqrt(1-(-12/13)^2)cos(s)=±1(1213)2

cos(s) = +-sqrt(169/169-144/169)cos(s)=±169169144169

cos(s) = +-sqrt(25/169)cos(s)=±25169

cos(s) = +-5/13" [2]"cos(s)=±513 [2]

We are told that ss is in the fourth quadrant, therefore, we shall choose the positive value:

cos(s) = 5/13cos(s)=513

Substitute sin(t)= 4/5sin(t)=45 into equation [1]:

cos(t) = +-sqrt(1-(4/5)^2)cos(t)=±1(45)2

cos(t) = +-sqrt(25/25-16/25)cos(t)=±25251625

cos(t) = +-sqrt(9/25)cos(t)=±925

cos(t) = +-3/5cos(t)=±35

We are told that tt is in the second quadrant, therefore, we shall choose the negative value:

cos(t) = -3/5" [3]"cos(t)=35 [3]

Using the identity,

cos(s+t) = cos(s)cos(t) - sin(s)sin(t)cos(s+t)=cos(s)cos(t)sin(s)sin(t)

, we substitute cos(s) = 5/13, cos(t) = -3/5, sin(s) = -12/13, and sin(t) = 4/5cos(s)=513,cos(t)=35,sin(s)=1213,andsin(t)=45:

cos(s+t) = (5/13)(-3/5) - (-12/13)(4/5)cos(s+t)=(513)(35)(1213)(45)

cos(s+t) = 33/65cos(s+t)=3365

Using the identity,

cos(s-t) = cos(s)cos(t) + sin(s)sin(t)cos(st)=cos(s)cos(t)+sin(s)sin(t)

, we substitute cos(s) = 5/13, cos(t) = -3/5, sin(s) = -12/13, and sin(t) = 4/5cos(s)=513,cos(t)=35,sin(s)=1213,andsin(t)=45:

cos(s-t) = (5/13)(-3/5) + (-12/13)(4/5)cos(st)=(513)(35)+(1213)(45)

cos(s-t) = -63/65cos(st)=6365