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=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−(45)2 and sinB=√1−(513)2sinB=√1−(513)2 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169⇒cosA=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=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1(45)2 and sin B=√1−(513)2 ⇒cosA=√1−1615  sin B=√1−25169  ⇒cosA=√25−1625 and sin B=√169−25169 ⇒cosA=√925 and sin B=√144169  cosA=35 and  sinB=1213  Now, cos(A+B)=cosA cosB−sinA sinB =35×513−45×1213 =1565−4865 =15−4865=−3365 (iii) We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−(45)2 and sinB=√1−(513)2 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169 ⇒cosA=35 and sinB=1213 Now, sin(A−B)=sinA cosB−cosA sinB =45×513−35×1213 =2065+3665=20−3665=−1665 (iv) We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−(45)2 sinB=√1−(513)2 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=169−25169 ⇒cosA=√925 and sinB=1213 Now, cos(A−B)=cosA cosB+sinA sinB =35×513+45×1213 =1565+4865 =15+4865 =6365

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