1
Question
Prove
that: cos 6x = 32 cos6 x – 48 cos4
x + 18 cos2 x – 1
Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
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Solution
L.H.S. =
cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos 2x [cos
3A = 4 cos3 A – 3 cos A]
= 4 [(2 cos2 x – 1)3 – 3 (2
cos2 x – 1) [cos 2x = 2 cos2
x – 1]
= 4 [(2 cos2 x)3 – (1)3
– 3 (2 cos2 x)2 + 3 (2 cos2
x)] – 6cos2 x + 3
= 4 [8cos6x – 1 – 12 cos4x
+ 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x
+ 24 cos2 x – 6 cos2x + 3
= 32 cos6x – 48 cos4x + 18
cos2x – 1
= R.H.S.
L.H.S. = cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]
= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]
= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
= 32 cos6x – 48 cos4x + 18 cos2x – 1
= R.H.S.
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