[20分]急問!!!trigonometry identities questions

( 1 ) sin2x + sin 3x = 0

2 sin ½ ( 2x + 3x ) cos ½ ( 2x – 3x ) = 0

2 sin 5x / 2 cos x / 2 = 0

sin 5x / 2 = 0 or cos x / 2 = 0

5x / 2 = 180*n or x / 2 = 360* n ± 90*

x = 72*n or x = 720*n ± 180* where n is an integer



( 2 ) 3 cos 2x + 5 sin x = 4

3 ( 1 – 2 sin2x ) + 5 sin x = 4

3 – 6 sin2 x + 5 sin x = 4

6 sin2x – 5 sin x + 1 = 0

( 2 sin x – 1 )( 3 sin x – 1 ) = 0

sin x = 1 / 2 or sin x = 1 / 3

x = 180* n + ( - 1 )n ( 30* ) or x = 180* n + ( - 1 )n ( 19.5* ) ( cor. to 3 s.f. )

where n is an integer



( 3 ) To prove : sin x cos ( x – y ) – cos x sin ( x – y ) = sin y

L.H.S. = sin x cos ( x – y ) – cos x sin ( x – y )

= sin { x – ( x – y )}

= sin y

= R.H.S.


2007-07-26 12:25:09 補充：
For Q3, we can directly apply sin ( A - B ) = sin A cos B - sin B cos A to get the answer.

(1)sin2x+sin3x=0
2sin(5x/2)cos(x/2)=0 (By using sum to product formula)
sin(5x/2)=0 or cos(x/2)=0
5x/2=180n(pi) or x/2=360n(pi)-90
∴x=72n(pi) or 720n(pi)-180

(2)3cos2x+5sinx=4
3(1-2sin^2x)+5sinx=4
6sin^2x-5sinx+1=0
(2sinx-1)(3sinx-1)=0
sinx=1/2 or 1/3
x=180n(pi)+-(-1)^n(30) or 180n(pi)+-(-1)^n(sin-1(1/3))

(3)LHS=sinxcos(x-y)-cosxsin(x-y)
=1/2[sin(2x-y)+siny]-1/2[sin(2x-y)+sin(-y)] (By using product to sum formula)
=1/2siny-1/2sin(-y)
=1/2siny+1/2siny
=siny
=RHS