A geometry problem by A Former Brilliant Member
If sin θ + sin 2 θ + sin 3 θ = 1 , then find the value of cos 6 θ − 4 cos 4 θ + 8 cos 2 θ .
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3 solutions
Given that sin 3 θ + sin 2 θ + sin θ = 1
cos 6 θ − 4 cos 4 θ + 8 cos 2 θ = ( 1 − sin 2 θ ) 3 − 4 ( 1 − sin 2 θ ) 2 + 8 ( 1 − sin 2 θ ) = 1 − 3 sin 2 θ + 3 sin 4 θ − sin 6 θ − 4 ( 1 − 2 sin 2 θ + sin 4 θ ) + 8 − 8 sin 2 θ = − sin 6 θ − sin 4 θ − 3 sin 2 θ + 5 = − sin 6 θ − sin 5 θ − sin 4 θ + sin 5 θ + sin 4 θ + sin 3 θ − sin 4 θ − sin 3 θ − sin 2 θ − 2 sin 2 θ + 5 = − sin 3 θ ( sin 3 θ + sin 2 θ + sin θ ) + sin 2 θ ( sin 3 θ + sin 2 θ + sin θ ) − sin θ ( sin 3 θ + sin 2 θ + sin θ ) − 2 sin 2 θ + 5 = − sin 3 θ + sin 2 θ − sin θ − 2 sin 2 θ + 5 = − sin 3 θ − sin 2 θ − sin θ + 5 = − ( sin 3 θ + sin 2 θ + sin θ ) + 5 = − 1 + 5 = 4
x+y+z=9, x 2+y 2+z 2=29, x 3+y 3+z 3=99 find x,y,z
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+
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=
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⟹
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S
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=
1
−
S
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.
B
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1
−
S
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=
C
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∴
(
1
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1
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(
4
−
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C
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C
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=
C
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.
⟹
4
−
4
C
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C
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4
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C
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2
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C
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6
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.
4
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6
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C
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.
I have used X in place of
θ
for ease of typing. The answer is
4
.
sin θ + sin 3 θ = 1 − sin 2 θ = cos 2 θ ⟹ sin θ ( 1 + 1 − cos 2 θ sin 2 θ ) = cos 2 θ Squaring both sides:
1 − cos 2 θ sin 2 θ ( 2 − cos 2 θ ) 2 = cos 4 θ
( 1 − cos 2 θ ) ( cos 4 θ − 4 cos 2 θ + 4 ) = cos 4 θ
⟹ cos 6 θ − 4 cos 4 θ + 8 cos 2 θ = 4