求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值 化简:(sinθ+sin3θ+sin5θ)/
问题描述:
求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值 化简:(sinθ+sin3θ+sin5θ)/
化简:(sinθ+sin3θ+sin5θ)/(cosθ+cos3θ+cos5θ)=
求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值
答
1.sin²αsin²β+cos²αcos²β-1/2cos2αcos2β=(1-cos^2a)(1-cos^2b)+cos^2a*cos^2b-1/2cos2αcos2β=1-cos^2a-cos^2b+2cos^2a*cos^2b-1/2cos2αcos2β=1-(1+cos2a)/2-(1+cos2b)/2+2*(1+cos2a)/...