急 會考程度amaths

1.a. sin3xcos4x = sin(x + 2x) cos4x

= [sinxcos2x + cosxsin2x]cos4x

= [sinxcos2x + cosx(2sinxcosx)]cos4x

= sinx [cos2x + 2cos^2x]cos4x

= sinx (cos2x + (1 + cos2x))cos4x

= sinx (1 + 2cos2x)cos4x

= sinx [cos4x + 2[1/2 (cos(4x - 2x) + cos(4x + 2x))]]

= sinx (cos4x + cos2x + cos6x)

= R.H.S.


b. From (a), put x = 15*

So, sin3(15*)cos4(15*) = sin15* [cos2(15*) + cos4(15*) + cos6(15*)]

sin45*cos60* = sin15* (cos30* + cos60* + cos90*)

1/2√2 = sin15* (√3/2 + 1/2 + 0)

sin15* = 1/2√2 [1/(√3/2 + 1/2)]

= 1/√2[√3 + 1]

= 1/(√6 + √2)

1.(a) L.H.S = (sin3x)(cos4x)
R.H.S = sinx(cos2x + cos4x + cos6x)
= sinx ( 2cos4x cos2x + cos 4x )
= cos4x ( sinx)(2cos2x + 1)
= cos4x ( 2sinxcos2x + sinx )
= cos4x ( sin 3x - sinx + sinx )
= (cos4x) (sin3x)
= L.H.S.
(b) By substituting x = 15°
sin 45° cos 60° = sin15°(cos30° + cos 60° + cos90°)
(√2/2)(1/2) = sin15°(√3/2 + 1/2)
sin15°= √2/[2(√3+ 1)]
= (√2)^2 / [ 2(√6 + √2) ]
= 1 / (√6 + √2)

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