1simply karlo naa...
a-b = cos^-1 1
a+b = cos^-1 (1/e)
etc. etc.
the number of ordered pairs (a,b) where a,b belongs to (-Ï€,Ï€) satisfying cos (a-b) =1 and cos (a+b) =1/e will be ???????????
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12 Answers
1a-b = 0 (a-b cant be taken 2pi ,,,, then a has to greater than 2pi which is not possible)
and a+b = cos^-1 (1/e) and 2pi-cos^-1 (1/e) both
then we get two pairs.
1357yeah it will be 4
a=b
so cos(2a)=1/e
2a=2nπ±cos-1(1/e)
a=nπ±(1/2)cos-1(1/e)
so a=±(1/2)cos-1(1/e), π-(1/2)cos-1(1/e), -π+(1/2)cos-1(1/e)