cosx+sinx=√2 sin3x=?
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Cos x + Sin x = √2
square on both sides:
Cos² x + Sin² x + 2 Sin x Cos x = 2
=> Sin 2x = 1
=> Cos 2x = √[1 - Sin² 2x ] = 0
(Cos x - Sin x)² = Cos² x + Sin² x - sin 2x = 1 - 1 = 0
=> Cos x = Sin x
=> Cos² x = Sin² x = 1/2
=> Cos x = Sin x = 1/√2
Sin 3 x = Sin x Cos 2x + Cos x Sin 2x
= 0 + 1/√2 * 1 = 1/√2
OR, alternately,
we know Sin 2x = 1 => x = π/4
=> Sin 3x = Sin 3π/4 =1/√2
square on both sides:
Cos² x + Sin² x + 2 Sin x Cos x = 2
=> Sin 2x = 1
=> Cos 2x = √[1 - Sin² 2x ] = 0
(Cos x - Sin x)² = Cos² x + Sin² x - sin 2x = 1 - 1 = 0
=> Cos x = Sin x
=> Cos² x = Sin² x = 1/2
=> Cos x = Sin x = 1/√2
Sin 3 x = Sin x Cos 2x + Cos x Sin 2x
= 0 + 1/√2 * 1 = 1/√2
OR, alternately,
we know Sin 2x = 1 => x = π/4
=> Sin 3x = Sin 3π/4 =1/√2
kvnmurty:
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