Question

if cos(A-B)+1=0 then prove cosA+cosB=0 and sinA+sinB=0

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Answer 1

Step-by-step explanation:

If cos(A+B) +1=0, then how to prove that cosA+ cos B=0 and sinA+sin B=0?

It is given that:

cos(A−B)+1=0

⇒cos(A−B)=−1.(1)

We also have:

(sinA+sinB)2+(cosA+cosB)2=sin2A+sin2B+2sinAsinB+cos2A+cos2B+2cosAcosB=1+1+2(cosAcosB+sinAsinB)[∵sin2θ+cos2θ=1,∀θ∈R.]=2+2cos(A−B)=2−2[Using (1).]

∴(sinA+sinB)2+(cosA+cosB)2=0.

But sum of two(real) perfect squares can be 0 iff both of the real numbers are separately 0. Therefore,

sinA+sinB=0and cosA+cosB=0.†