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What is the value of sin18° |
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Answer» √5-1/4 Let θ=18°5θ=90°2θ+3θ=90°Sin2θ=sin(90°-3θ)Sin2θ=cos3θ2sinθcosθ=4cos(cube)θ-3cosθ2sinθ=4cos(square)θ-32sinθ=4(1-sin(square)θ)-34sin(square)θ+2sinθ-1=0(Sinθ=sin18°)Sin18°=-2+_under root 4+16÷2×4Sin18°=under root 5-1÷4 Can not be able to find |
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