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Question
If 2 sin x = `sqrt3` , evaluate.
(i) 4 sin3 x - 3 sin x.
(ii) 3 cos x - 4 cos3 x.
Solution
2 sin x = `sqrt3`
sin x = `sqrt3 /(2)`
i.e.`"perpendicular"/"base" = "BC"/"AC" = sqrt3/(2)`
Therefore if length of perpendicular = `sqrt3x` , length of = 2x
Since
AB2 + BC2 = AC2 ...[ Using Pythagoras Theorem]
(2x)2 – (`sqrt3x`)2 = AB2
AB2 = x2
∴ AB = x
Now, cos x = `"AB"/"AC" = (1)/(2)`
(i) 4 sin3 x – 3sin x
= `4 (sqrt3/2)^3 – 3(sqrt3/2)`
= `(3sqrt3)/2 – (3sqrt3)/2`
= 0
(ii) 3 cos x – 4 cos3 x
= `3 * (1)/(2) – 4 * (1/2)^3`
= `3 * 1/2 - 4 * 1/8`
= `3 * 1/2 - cancel4^1 * 1/cancel8_2`
= `3/2 – 1/2`
= 1
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