# Harder Trig identity

[mentor note] moved to homework forum hence no template.

HI,

I'm just having a bit of trouble with the numerator part of this identity ...

Resolving the the denominator is fairly straightforward but ..

Can anyone shed light on the final couple of steps

(sin^3 x - cos^3 x)/(sinx + cosx) = (csc^2 x - cotx - 2cos^2 x)/(1 - cot^2 x)

(sine cubed, cosine cubed, cosec squared, cosine squared, cotangent squared)

I hope I've transcribed it correctly .

Peter

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## Answers and Replies

for the LHS i used a multiplier/simplifier of (sinx - cosx)/(sin^2 x) and re-arrangement to achieve the denominator of the RHS
P

jedishrfu
Mentor
Where is your work?

I don't see any steps to solving it only the trig identity itself or is that where you are stuck.

Please be aware that we can't do your homework for you but can only provide hints once we see your work.

Lastly, have you noticed that LHS = s^3 - c^3 = (s - c) (s^2 + sc + c^2) = (s - c) * sc ?

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SammyS
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for the LHS i used a multiplier/simplifier of (sinx - cosx)/(sin^2 x) and re-arrangement to achieve the denominator of the RHS
P
Where does anything like (sinx - cosx)/(sin2 x) appear on the left hand side ?

sorry sammy .. I meant top and bottom multiplied by that because it eliminates stuff and gives the denominator straight away but leaves the numerator slightly more complicated ...

It's not homework guys , just a challenge ................. and the theory is that we learn a bit plus a bit of experience along the way ..

Jedishrfu ..... yes I got that ... and used the fraction top and bottom to get the RHS denominator right away .. but the top is proving to be a pig .. unless there's something dead easy I'm missing

but tbh, Jedishrfu, I thought it was more like (s-c)* (1+ sc)

SammyS
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but tbh, Jedishrfu, I thought it was more like (s-c)* (1+ sc)
That is correct.

And you're multiplying that by another (s - c) . Right?
... then dividing by s2 .

The result follows quite directly after that.

yessss, it does ... thanks for that :-)