if sinA + sin B = a , cos A + cos B = b then prove that cos ( A+ B) = b^- a^/b^-a^
Answers
Answer:
a=sinA+sinB
a2=sin2A+sin2B+2sinAsinB
b=cosA+cosB
b2=cos2A+cos2B+2cosAcosB
b2+a2=2+2(cosAcosB+sinAsinB)=2+2cos(A−B)
cos(A−B)=b2+a2−22
b2−a2
=cos2A−sin2A+cos2B−sin2B+2(cosAcosB−sinAsinB)
=cos2A+cos2B+2cos(A+B)
There’s a formula for cos2A+cos2B called the beating formula. Let’s derive it quickly:
cos(x+y)=cosxcosy−sinxsiny
cos(x−y)=cosxcosy+sinxsiny
cos(x+y)+cos(x−y)=2cosxcosy
Let 2A=x+y,2B=x−y so x=A+B,y=A−B.
cos2A+cos2B=2cos(A+B)cos(A−B)
OK, let’s plug that in and cos(A−B)=b2+a2−22 in and see where we are:
b2−a2=2cos(A+B)cos(A−B)+2cos(A+B)
=2(1+cos(A−B))cos(A+B)
=2(22+b2+a2−22)cos(A+B)
cos(A+B)=b2−a2b2+a2
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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