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If x+y+z=pi/2, then prove that |sinxsiny...

If `x+y+z=pi/2,` then prove that `|sinxsinysinzcosxcosycoszcos^3xcos^y ycos^3z|=0`

Text Solution

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D=`|{:(sinx,siny,sinz),(cosx,cosy,cosz),(cos^(3)x,cos^(3)y,cos^(3)z):}|`
Expanding along `R_(3)`, we get
`D=sum cos^(3)x sin(y-z)`
Given `x+y+z=(pi)/(2)`
`therefore D=sumsin^(3)(y+z)sin(y-z)`
`=sumsin^(2)(y+z)sin(y+z)sin(y-z)`
`=sum (1-sin^(2)x)(sin^(2)y-sin^(2)z)`
=0
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