2xy
Q.30 If sinA+ sin B = x and cos A+ cos B=y then show that sin(A + B) = 2xy/x²+y²
Answers
Answer:
It is given that A+B+C=0⇒C=−(A+B)
Now, x
2
+2xysinC+y
2
=sin
2
A+2sinAcosBsinC+cos
2
B=sin
2
A+2sinAcosBsin(−(A+B))+cos
2
B
=sin
2
A−2sinAcosB[sinAcosB+cosAsinB]+cos
2
B
=sin
2
A−2sinAcosBsinAcosB−2sinAcosBcosAsinB+cos
2
B
=sin
2
A−sin
2
Acos
2
B+cos
2
B−sin
2
Acos
2
B−2sinAcosBcosAsinB
=sin
2
A(1−cos
2
B)+cos
2
B(1−sin
2
A)−2sinAcosBcosAsinB
=sin
2
Asin
2
B+cos
2
Bcos
2
A−2sinAcosBcosAsinB
=sin
2
Asin
2
B−sinAcosBcosAsinB+cos
2
Bcos
2
A−sinAcosBcosAsinB
=sinAsinB[sinAsinB−cosAcosB]+cosAcosB[cosAcosB−sinAsinB]
=sinAsinB[sinAsinB−cosAcosB]−cosAcosB[sinAsinB−cosAcosB]
=[sinAsinB−cosAcosB][sinAsinB−cosAcosB]
=[−cos(A+B)][−cos(A+B)]
which can be written as
=cos(−(A+B))cos(−(A+B))=cos
2
Step-by-step explanation:
prove that
Sin(A+B)= 2ab/ a^2+b^2
Sin(A+B) = sin A cos B +CosA SinB
Cos(A+B) = cos A cosB + SinA SinB
Finding a^2 +b^2 (1)
a^2 +b^2= sin^2A +sin^2B +2sinA SinB + Cos^2A +Cos^2B + 2Cos A Cos B
= 1+1 +2 ( CosA CosB + SinA SinB )
=2+2( CosA cos B +sinA SinB)
a^2+b^2 --2 =2 Cos(A-B) )
Cos (A-B) = (a^2+b^2 )/2 --1 (1)
Finding values of ab (2)
2ab =2 (sinA +SinB ) (CosA +CosB)
= 2 [ sinA CosA + SinA cos B + SinB Cos A + SinB Cos B]
2ab= 2 (1+ Cos A CosB + SinA SinB )
ab= [1+ cos(A --B)]
ab=1 +(a^2+b^2)/2 --1 from (1)
ab=( a^2+b^2)/2
Hence sin(A+B) = 2ab /a^2+b^2
LH S =R H S