已知sin(x/2)=2cos(x/2),求(cos2x)÷[﹙√2﹚cos(45°+x)]sinx
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已知sin(x/2)=2cos(x/2),求(cos2x)÷[﹙√2﹚cos(45°+x)]sinx
∵sin(x/2)=2cos(x/2)
∴sin(x/2)/cos(x/2)=2
cos(x/2)/sin(x/2)=1/2
∴(cos2x)÷{[﹙√2﹚cos(45°+x)]sinx}
=(cos²x-sin²x)÷{[√2(cos45ºcosx-sin45ºsinx)]sinx}
=(cosx+sinx)(cosx-sinx)/[cosx-sinx)sinx]
=(cosx+sinx)/sinx
=cosx/sinx+1
=[cos²(x/2)-sin²(x/2)]/(2sinx/2cosx/2)+1
=cos(x/2)/(2sinx/2)-sin(x/2)/(2cosx)+1
=1/4-1+1
=1/4
祝进步!
∴sin(x/2)/cos(x/2)=2
cos(x/2)/sin(x/2)=1/2
∴(cos2x)÷{[﹙√2﹚cos(45°+x)]sinx}
=(cos²x-sin²x)÷{[√2(cos45ºcosx-sin45ºsinx)]sinx}
=(cosx+sinx)(cosx-sinx)/[cosx-sinx)sinx]
=(cosx+sinx)/sinx
=cosx/sinx+1
=[cos²(x/2)-sin²(x/2)]/(2sinx/2cosx/2)+1
=cos(x/2)/(2sinx/2)-sin(x/2)/(2cosx)+1
=1/4-1+1
=1/4
祝进步!
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