sinθ-0.2cosθ等于0.5 θ等于多少
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sinθ-0.2cosθ等于0.5 θ等于多少
郭敦顒回答:
sinθ-0.2cosθ=0.5
用试算—逐步逼近法求解.
当 θ =45°时,sinθ-0.2cosθ=0.7071-0.2×0.7071=0.5657,
误差=0.5657-0.5=0.0657;
设θ=43°,则sinθ-0.2cosθ=0.682-0.2×0.73135=0.5357,
误差=0.5357-0.5=0.0357;
设θ=41°,则sinθ-0.2cosθ=0.6561-0.2×0.77547=0.5052,
误差=0.5052-0.5=0.0052,相对误差=0 .0052/5×100%=0.104%;
设θ=40 .6°,
则sinθ-0.2cosθ=0.6508-0.2×0.75927=0.6508-0 .1518=0 .499
误差=0.499-0.5=-0.001,相对误差=-0 .001/5×100%=-0.02%
设θ=40 .75°,
则sinθ-0.2cosθ=0.65276-0.2×0.75756=0.65276-0 .15151=0 .50125
误差=0.50125-0.5=0.00125,相对误差=0 .00125/5×100%=0.025%;
设θ=40 .67°,
则sinθ-0.2cosθ=0.6517-0.2×0.75848=0.6517-0 .1517=0 .5,
误差=0.5-0.5=0.
∴θ=40.67°.
sinθ-0.2cosθ=0.5
用试算—逐步逼近法求解.
当 θ =45°时,sinθ-0.2cosθ=0.7071-0.2×0.7071=0.5657,
误差=0.5657-0.5=0.0657;
设θ=43°,则sinθ-0.2cosθ=0.682-0.2×0.73135=0.5357,
误差=0.5357-0.5=0.0357;
设θ=41°,则sinθ-0.2cosθ=0.6561-0.2×0.77547=0.5052,
误差=0.5052-0.5=0.0052,相对误差=0 .0052/5×100%=0.104%;
设θ=40 .6°,
则sinθ-0.2cosθ=0.6508-0.2×0.75927=0.6508-0 .1518=0 .499
误差=0.499-0.5=-0.001,相对误差=-0 .001/5×100%=-0.02%
设θ=40 .75°,
则sinθ-0.2cosθ=0.65276-0.2×0.75756=0.65276-0 .15151=0 .50125
误差=0.50125-0.5=0.00125,相对误差=0 .00125/5×100%=0.025%;
设θ=40 .67°,
则sinθ-0.2cosθ=0.6517-0.2×0.75848=0.6517-0 .1517=0 .5,
误差=0.5-0.5=0.
∴θ=40.67°.
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