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Question

If cosA=2425 and cosB=35, where π<A<3π2 and 3π2<B<2π, find the following:
(i)sin(A+B) (ii)cos(A+B)

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Solution

We have,
cosA=2425 and cosB=35
sinA=1cos2A
sinB=1cos2B
[ In the 3rd and 4th quadrant sinθ is negative]
sinA=1(2425)2 and
sinB=1(35)2
sinA=1576625 and
sinB=925
sinA=49625 and
sinB=1625
sinA=725 and sinB=45
Now,
(i) sin(A+B)=sinAcosB+cosAsinB
=725×352425(45)
=2125+96125
=75125=35
(ii) cos(A+B)=cosAcosBsinAsinB
=2425×35(725)×(45)
=7212528125
=7228125=100125=45


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