Andy C.
asked • 04/04/22Trig Question I cant figure out how to solve!
If cosα=0.558 and cosβ=0.858 with both angles’ terminal rays in Quadrant-I, find the values of(a)cos(α+β)= (b)cos(α−β)=
(a)
cos(α+β)=
(b)
cos(α−β)=
More
1 Expert Answer
Raymond B. answered • 04/04/22
Tutor
5
(2)
Math, microeconomics or criminal justice
cos(A+B) = cosAcosB - sinAsinB = .558(.858) - .830(.514)= .479- .426 = .053
cos(A-B) = cosAcosB + sinAsinB = .479 + .426 = .905
A= about 56.1 degrees
B = about 30.9
A+B = about 87 degrees
A-B = about 25.2 degrees
cos(87) = .052
cos(25.2) = .905
use an online calculator such a Desmos or use a handheld calculator with inverse trig functions
if cosA = .558, then sinA = sqr(1- cos^2(A) = sqr(1-.558^2) = sqr(1 -.311) = sqr.689 = .830
solve for sinB the same way
then plug those values into the sum and difference formulas for cosines
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Mark M.
These are calculations not equations to be solved. Use the appropriate formulas.04/04/22