Prove this identity without using cross multiplication by manipulating one side using trig identitie

taoigas

taoigas

Answered question

2022-01-29

Prove this identity without using cross multiplication by manipulating one side using trig identities:
sin3xcos3xsinx+cosx=csc2xcotx2cos2x1cot2x
I first started off on the LHS and managed to get the denominator to become 1cot2x by multiplying by sinxcosx and then dividing by sin2x, but from there I had no idea how to continue.

Answer & Explanation

trovabile4p

trovabile4p

Beginner2022-01-30Added 13 answers

Let s=sinx and c=cosx
s3c3s+cscsc=s4+c4sc(s2+c2)s2c2
Now s4+c4=(s2+c2)22s2c2 and divide top & bottom by s2
1s2cs2c21c2s2
spelkw

spelkw

Beginner2022-01-31Added 12 answers

Note that
sin3(x)cos3(x)=(sin(x)cos(x))(sin2(x)+cos2(x)+sin(x)cos(x))
=(sin(x)cos(x))(1+sin(x)cos(x))
Therefore,
sin3(x)cos3(x)sin(x)+cos(x)=(sin(x)cos(x))(1+sin(x)cos(x))sin(x)+cos(x)
=(sin(x)cos(x))2(1+sin(x)cos(x))sin2(x)cos2(x)
=(12sin(x)cos(x))(1+sin(x)cos(x))sin2(x)cos2(x)
=1sin(x)cos(x)2sin2(x)cos2(x)sin2(x)cos2(x)
=sin2(x)sin2(x)csc2(x)cot(x)2cos2(x)1cot2(x)

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