cos³θ+cos³(2π/3+θ)+cos³(4π/3+θ)= acos3θ,then 4a=
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1
Answer:
3/4
Step-by-step explanation:
cos3x=4cos
3
x−3cosx
⟹cos
3
x=
4
cos3x+3cosx
cos
3
θ+cos
3
(
3
2π
+θ)+cos
3
(
3
4π
+θ)
=
4
cos3θ+3cosθ
+
4
cos3(
3
2π
+θ)+3cos(
3
2π
+θ)
+
4
cos3(
3
4π
+θ)+3cos(
3
4π
+θ)
=
4
cos3θ+3cosθ
+
4
cos(2π+3θ)+3cos(
3
2π
+θ)
+
4
cos(4π+3θ)+3cos(
3
4π
+θ)
=
4
cos3θ+3cosθ
+
4
cos(3θ)+3cos(
3
2π
+θ)
+
4
cos(3θ)+3cos(
3
4π
+θ)
=
4
3cos3θ+3cosθ
+
4
3(cos(
3
2π
+θ)+cos(
3
4π
+θ))
=
4
3cos3θ+3cosθ
+
4
3(2cos(π+θ)cos(
3
−π
))
=
4
3cos3θ+3cosθ
+
4
3cos(π+θ)
=
4
3cos3θ+3cosθ
−
4
3cos(θ)
=
4
3cos3θ
⟹a=
4
3
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