Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To solve $\sin^3x+\cos^3x=1$

So I just thought of a solution like:

Let $\sin x=t$. Then we have: $$t^3+(1-t^2) \sqrt{1-t^2} =1 \\ (1-t^2) \sqrt{1-t^2} =1-t^3 \\ (1-t^2)^3=(1-t^3)^2 \\ (1-t)^3(1+t)^3=(1-t)^2(1+t+t^2)^2 \\ (1-t)^2\left[ (1-t)(1+t)^3-(1+t+t^2)^2 \right]=0$$

Which is followed by $\sin x=0$ or $\sin x=1$. However, this solution seems very easy to make a silly mistake in (with all the squares and cubes of differences and sums). Is there any easier solution to this?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Note that $\sin x$ and $\cos x$ are always at most $1$, so $\sin^3 x+\cos^3 x\leq \sin^2 x+\cos^2 x=1$ with equality only $\sin x,\cos x$ are $0$ or $1$.

share|improve this answer
    
Hmm.. how do we get to $sin x = 0$ or $sin x = 1$ just from $sin^3x+cos^3x\leq1$? –  Straightfw Apr 30 '12 at 18:05
2  
Hint: If $\sin x<1$ and $\sin x\neq 0$ then $\sin^3 x < \sin^2 x$ @Straightfw –  Thomas Andrews Apr 30 '12 at 18:09
    
Right! OHMIGOSH. Thank you very much, that sollution is brilliant :D –  Straightfw Apr 30 '12 at 18:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.