How do you simplify #cot(x+y)+cos(4x-y)# to trigonometric functions of x and y?

Question

Samantha Abate · Accepted Answer

#cot(x+y)+cos(4x-y)#

=#(cotxcoty+1)/(coty-cotx)+cosy(8cos^4x-8cos^2x+1)+siny(8sinxcos^3x-4sinxcosx)# We can use the identities #cos(A-B)=cosAcosB+sinAsinB# and #cot(A+B)=(cotAcotB+1)/(cotB-cotA)# Hence #cot(x+y)+cos(4x-y)# = #(cotxcoty+1)/(coty-cotx)+cos4xcosy+sin4xsiny# And now using #cos2A=2cos^2A-1# and #sin2A=2sinAcosA# #cos4x=2cos^2(2x)-1=2(2cos^2x-1)^2-1=8cos^4x-8cos^2x+1# and #sin4x=2sin2xcos2x=4sinxcosx(2cos^2x-1)=8sinxcos^3x-4sinxcosx# Hence, #cot(x+y)+cos(4x-y)# = #(cotxcoty+1)/(coty-cotx)+cosy(8cos^4x-8cos^2x+1)+siny(8sinxcos^3x-4sinxcosx)#

Zoe Austin · Answer

undefined To simplify cot(x+y) + cos(4x-y) to trigonometric functions of x and y, we can use trigonometric identities. One approach is to express cot(x+y) and cos(4x-y) in terms of sine and cosine functions using the identities:

1. cot(x+y) = cos(x+y)/sin(x+y)
2. cos(4x-y) = cos(4x)cos(y) + sin(4x)sin(y)

Then, substitute these expressions into the original equation and simplify further by expanding and combining terms using trigonometric identities until the expression is in terms of sine and cosine functions of x and y.