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If in a triangle ABC, ab = c2  prove that

                cos(A - B) + cos C + cos 2C = 1

 

NB - could only solve if a and b are equal to c [equilateal]...but not sure if correct reasoning.

Hope you are having a good day...regards 

 May 26, 2019
edited by OldTimer  May 26, 2019
edited by OldTimer  May 26, 2019
 #1
avatar+397 
+5

From the sine rule asin(A)=bsin(B)=csin(C),

we can say that 

absin(A)sin(B)=c2sin2(C),

and since we are told that ab=c2,

it follows that 

sin(A)sin(B)=sin2(C)(1)

 

Expanding the LHS of the (supposed) identity using standard trig identities, we have

 

cos(A)cos(B)+sin(A)sin(B)+cos(C)+12sin2(C)

 

which equals, (making use of (1)),

 

cos(A)cos(B)sin(A)sin(B)+cos(C)+1

=cos(A+B)+cos(C)+1

=cos(180C)+cos(C)+1

=cos(C)+cos(C)+1

=1.

 May 27, 2019
 #2
avatar+239 
+1

Thanks that was great Tiggsy!!!

 May 28, 2019

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