Asked by sankalp
Show that (cosð + cosß)² + (sinð + sinß)² = 4cot(ð-ß/2)
Answers
Answered by
Reiny
I will write it as:
(cosx + cosy)² + (sinx + siny)² = 4cot(x-y/2) --> easier to type
LS = cos^2 x + 2cosxcosy + cos^2 y + sin^2 x + 2sinxsiny + sin^2 x
= sin^2 x + cos^2 x + sin^ y + cos^2 y + 2(cosxcosy + sinxsiny)
= 2 + 2cos(x - y)
= 2(1 + cos(x-y) )
RS = 4 cot(x - y/2)
= 4(1 + tanxtan y/2)/(tanx - tan y/2)
= ........
see what you can do with that.
(cosx + cosy)² + (sinx + siny)² = 4cot(x-y/2) --> easier to type
LS = cos^2 x + 2cosxcosy + cos^2 y + sin^2 x + 2sinxsiny + sin^2 x
= sin^2 x + cos^2 x + sin^ y + cos^2 y + 2(cosxcosy + sinxsiny)
= 2 + 2cos(x - y)
= 2(1 + cos(x-y) )
RS = 4 cot(x - y/2)
= 4(1 + tanxtan y/2)/(tanx - tan y/2)
= ........
see what you can do with that.
Answered by
Steve
If that fails, try cot (x-y)/2
and use your half-angle formula, and more parentheses next time.
and use your half-angle formula, and more parentheses next time.
Answered by
Reiny
ahh, the old "missing bracket" trap.
Thanks Steve.
Thanks Steve.
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