Prove that sinA/2sin7A/2+sin3A/2sin11A/2=sin2Asin5A

Asked by Anil | 6th Jun, 2017, 05:52: PM

Expert Answer:

begin mathsize 16px style sin straight A over 2 sin fraction numerator 7 straight A over denominator 2 end fraction plus sin fraction numerator 3 straight A over denominator 2 end fraction sin fraction numerator 11 straight A over denominator 2 end fraction
equals 1 half open parentheses 2 sin fraction numerator begin display style straight A end style over denominator begin display style 2 end style end fraction sin fraction numerator begin display style 7 straight A end style over denominator begin display style 2 end style end fraction plus 2 sin fraction numerator begin display style 3 straight A end style over denominator begin display style 2 end style end fraction sin fraction numerator begin display style 11 straight A end style over denominator begin display style 2 end style end fraction close parentheses
Using space 2 sinAsinB equals cos open parentheses straight A minus straight B close parentheses minus cos open parentheses straight A plus straight B close parentheses
equals 1 half open parentheses cos 3 straight A minus cos 4 straight A plus cos 4 straight A minus cos 7 straight A close parentheses
equals fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction open parentheses cos 3 straight A minus cos 7 straight A close parentheses
equals 1 half cross times 2 sin 5 straight A space sin 2 straight A
equals sin 5 straight A space sin 2 straight A
end style

Answered by Sneha shidid | 7th Jun, 2017, 08:51: AM