Question

PROVE THAT cos(a + b) cos x - cos(b + x) sin a = sinb sin (x - a)

Answer 1

Answer:

If cosx+cosy+cosz=0

sinx+siny+sinz=0

Show that cos(x−y)=cos(y−z)=cos(z−x)=

2

−3

∵cosx+cosy=−cosz

or, cos

2

x+cos

2

y+2cosx.cosy=cos

2

z

sinx+siny=−sinz

sin

2

x+sin

2

y+2sinx.siny=sin

2

z

Adding both side we get,

(cos

2

x+sin

2

x)+(cos

2

y+sin

2

y)+2(cosx.cosy+sinx.siny)=cos

2

z+sin

2

z

or, 1+1+2cos(x−y)=1

or, cos−(x−y)=

2

−1

Similarly we can prove,

sinx+sinz=siny

⟹sin

2

x+sin

2

z+2sinx.sinz=sin

2

y

⟹cosx+cosz=−cosy

⟹cos

2

x+cos

2

z+2cosx.cosz=cos

2

y

adding both we get,

1+1+2cos(z−x)=1

cos(z−x)=

2

−1

similarly cos(y−z)=

2

−1

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