標題:
sin(θ-30°)=2cos(θ+60°)
發問:
Solve: sin(θ-30°)=2cos(θ+60°) Can I do it in this way? sin(θ-30°)=2cos(θ+60°) cos[90°-θ+30°]=2cos(θ+60°) cos(θ-120°)=2cos(θ+60°) -cos[180+θ-120°]=2cos(θ+60°) -cos(θ+60°)=2cos(θ+60°) 3cos(θ+60°)=0 cos(θ+60°)=0 θ+60°=360°n±90 θ=360°n+30° or θ=360°n-150° But my teacher's answer is θ=180°n+30°.....
最佳解答:
sin(θ-30°)=2cos(θ+60°) sin θ cos 30* - sin 30* cos θ = 2 cos θ cos 60* - 2 sin θ sin 60* √3/2 sin θ - 1/2 cos θ = 2(1/2) cos θ - 2 (√3/2 ) sin θ √3/2 sin θ - 1/2 cos θ = cos θ - √3 sin θ 3√3/2 sin θ = 3 / 2 cos θ √3 sin θ = cos θ tan θ = 1/ √3 θ = 180°n+30°where n is an integer
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其他解答:
sin(θ-30°)=2cos(θ+60°) cos[90°-θ+30°]=2cos(θ+60°) cos(θ-120°)=2cos(θ+60°) -cos[180+θ-120°]=2cos(θ+60°) -cos(θ+60°)=2cos(θ+60°) 3cos(θ+60°)=0 cos(θ+60°)=0 θ+60°=360°n±90° θ=360°n+30° or θ=360°n-150° where n is an integer. they are correct,but you can simplify it！！ let n be 1, θ=360°n+30° or θ=360°n-150° θ=360°(1)+30° or θ=360°(1)-150° θ=390° or θ=210° which is 180°(2)+30° or 180°(1)+30° Since n can be any integer,the general form will be θ=180°n+30° where n is an integer. you can also draw graphs yourself,then you will understand it!!!!!!!!!!!!!!!!!!!|||||sin(θ-30°)=2cos(θ+60°) sin[-90+(θ+60)]=2cos(θ+60) -sin[90-(θ+60)]=2cos(θ+60) -cos(θ+60)=2cos(θ+60) 3cos(θ+60)=0 cos(θ+60)=cos(90) cosθ=cos30
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