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here
cos(n+1)xcos(n+2)x + sin(n+1)xsin(n+2)x
it is from the identity
cos(a-b) = cosa * cosb + sina * sinb
here a= (n+1)x b= (n+2)x
so,
cos((n+1)x-(n+2)x)
cos(nx+x-nx-2x)
cos(-x)
cosx { since cos(-x) = cosx }
cos(n+1)xcos(n+2)x + sin(n+1)xsin(n+2)x
it is from the identity
cos(a-b) = cosa * cosb + sina * sinb
here a= (n+1)x b= (n+2)x
so,
cos((n+1)x-(n+2)x)
cos(nx+x-nx-2x)
cos(-x)
cosx { since cos(-x) = cosx }
Anonymous:
thanks ✌✌
Answered by
2
Hey!!!!
Good Afternoon
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Sorry for the untidy work
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Hope this helps ✌️
Good Afternoon
_________________
Refer to the attachment
Sorry for the untidy work
_________________
Hope this helps ✌️
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