the 2 one please....if Any solve it
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hence it is proved
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cos 60°=1/2
sin^2 30°=1/2
cos^2 30°=√3/2
so,
cos 60°=(1- 2sin^2 30°)
1/2={1 - 2*(1/2)^2 }
1/2=(1 - 2*1/4)
1/2=(1 - 1/2)
1/2=1/2 (proved)
cos 60°=(2*cos^2 30° - 1)
1/2={2*(√3/2)^2 - 1}
1/2={2*3/4 -1}
1/2=(3/2 - 1)
1/2=1/2 (proved)
sin^2 30°=1/2
cos^2 30°=√3/2
so,
cos 60°=(1- 2sin^2 30°)
1/2={1 - 2*(1/2)^2 }
1/2=(1 - 2*1/4)
1/2=(1 - 1/2)
1/2=1/2 (proved)
cos 60°=(2*cos^2 30° - 1)
1/2={2*(√3/2)^2 - 1}
1/2={2*3/4 -1}
1/2=(3/2 - 1)
1/2=1/2 (proved)
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