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Question Number 111125 by bemath last updated on 02/Sep/20

Answered by Dwaipayan Shikari last updated on 02/Sep/20

cos55°cos65°cos175° −(1/2)(2cos55°cos5°)cos65° −(1/2)(cos60°+cos50°)sin25° −(1/4)sin25°−(1/4)(2cos50°sin25°) =−(1/4)sin25°−(1/4)sin75°+(1/4)sin25° =−(1/4).cos15°=−(1/8)(√(2+(√3))) 1+cos30=2cos^2 15°⇒(√((1/2).((2+(√3))/2) ))=cos15°=(1/2)(√(2+(√3)))

cos55°cos65°cos175°−12(2cos55°cos5°)cos65°−12(cos60°+cos50°)sin25°−14sin25°−14(2cos50°sin25°)=−14sin25°−14sin75°+14sin25°=−14.cos15°=−182+31+cos30=2cos215°⇒12.2+32=cos15°=122+3

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