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Question

If sinA=45 and cosB=513, where 0<A, B<π2, find the values of the following:
(i) sin(A+B)
(ii) cos(A+B)
(iii) sin(A-B)
(iv) cos(A-B)


Solution

We have,
sinA=45 and B=513
cosA=1sin2A and sinB=1cos2B
cosA=1(45)2 and sinB=1(513)2sinB=1(513)2
cosA=11625 and sinB=125169
cosA=251625 and sinB=16925169
cosA=925 and sinB=144169cosA=35 and sinB=1213
Now,
sin(A+B)=sinA cosB+cosA sinB
=45×513+35×1213=2065+3665=20+3665=5665
(ii) We have,
sinA=45 and B=513
cosA=1sin2A and sinB=1cos2B
cosA=1(45)2 and
sin B=1(513)2
cosA=11615 
sin B=125169 
cosA=251625 and
sin B=16925169
cosA=925 and
sin B=144169 
cosA=35 and 
sinB=1213 
Now,
cos(A+B)=cosA cosBsinA sinB
=35×51345×1213
=15654865
=154865=3365
(iii) We have,
sinA=45 and cosB=513
cosA=1sin2A and sinB=1cos2B
cosA=1(45)2 and
sinB=1(513)2
cosA=11625 and sinB=125169
cosA=251625 and
sinB=16925169
cosA=925 and sinB=144169
cosA=35 and sinB=1213
Now,
sin(AB)=sinA cosBcosA sinB
=45×51335×1213
=2065+3665=203665=1665
(iv) We have,
sinA=45 and cosB=513
cosA=1sin2A and
sinB=1cos2B
cosA=1(45)2
sinB=1(513)2
cosA=11625 and
sinB=125169
cosA=251625 and
sinB=16925169
cosA=925 and sinB=1213
Now,
cos(AB)=cosA cosB+sinA sinB
=35×513+45×1213
=1565+4865
=15+4865
=6365

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