Trig. fun problem solving: Decide sin^3 x + cos^3 x in terms of/ if sin x + cos x = a

[cosine]

Decide sin^3 x + cos^3 x in terms of/ if sin x + cos x = a

Finding a is not necessary for this problem. The answer is a(3 - a^2)/ 2, problem is how do you get there? Thanks

 

[ksdhart2]

Well, first off I'll assume there was a minor typo and the problems asks you to rewrite the expression in terms of "a", rather than "/" If that's not right, please provide any necessary corrections.

Now, as for the problem itself, the initial given form looks exactly like a sum of cubes. What happens if you apply the usual formula for that, letting a = sin(x) and b = cos(x)? Where does that lead?

 

[pka]

Decide sin^3 x + cos^3 x in terms of/ if sin x + cos x = a

Finding a is not necessary for this problem. The answer is a(3 - a^2)/ 2, problem is how do you get there? Thanks

\(\displaystyle \begin{align*}\sin^3(x)+\cos^3(x)&=(\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x))\\&=a(1-\sin(x)\cos(x))\\&=a[3-(\sin(x)+\cos(x))^2]/2\end{align*}\)

 

[cosine]

\(\displaystyle \begin{align*}\sin^3(x)+\cos^3(x)&=(\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x))\\&=a(1-\sin(x)\cos(x))\\&=a[3-(\sin(x)+\cos(x))^2]/2\end{align*}\)

thank you, but I don't understand how you go from the second row to the last? Could you please explain, I would appreciate it, thanks

 

[pka]

\(\displaystyle \begin{align*}\sin^3(x)+\cos^3(x)&=(\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x))\\&=a(1-\sin(x)\cos(x))\\&=a[3-(\sin(x)+\cos(x))^2]/2\end{align*}\)

thank you, but I don't understand how you go from the second row to the last? Could you please explain, I would appreciate it, thanks

\(\displaystyle \begin{align*}a[3-(\sin(x)+\cos(x))^2]&=a[3-(\sin^2(x)+2\sin(x)\cos(x)+\cos(x)^2)] \\&=a[3-(1+2\sin(x)\cos(x)]\\&=a[2-2\sin(x)\cos(x)] \end{align*}\)