Let a be a real number such that 0 < a < \frac{\pi}{2}. Show that \left(\sin(a)\right)^7 + \left(\cos ( a)\right)^7< 2 sin(a)^4 cos(a)^4.

Answer 1

We know that for any real number x, we have 0 < \sin(x), \cos(x) < 1. Thus, let's take the expression on the right side of the inequality and factorize it as follows:

2(\sin(a)^4)(\cos(a)^4) = 2[(\sin(a)^2)(\cos(a)^2)]^2 = 2[(\sin(2a)/2)^2]^2 = \frac{1}{2^4}\sin(2a)^4.

So, now we have to prove that (\sin(a))^7 + (\cos (a))^7 < \frac{1}{2^4}\sin(2a)^4. Let's look at the expression on the left side:

(\sin(a))^7 + (\cos (a))^7 = (\sin(a) + \cos(a))(\sin^6(a) - \sin^5(a)\cos(a) + \sin^4(a)\cos^2(a) - \sin^3(a)\cos^3(a) + \sin^2(a)\cos^4(a) - \sin(a)\cos^5(a) + \cos^6(a)).

Now, we know that for 0 < a < \frac{\pi}{2}, both \sin(a) and \cos(a) are positive, making the expression \sin(a) + \cos(a) > 0.

Additionally, we know that
\begin{aligned}
\sin(2a) &= 2\sin(a)\cos(a)\\ \cos(2a) &= \cos^2(a) - \sin^2(a).
\end{aligned}

Squared, we get:
\begin{aligned}
\sin^2(2a) &= 4\sin^2(a)\cos^2(a)\\
\cos^2(2a) &= \cos^4(a) - 2\sin^2(a)\cos^2(a) + \sin^4(a).
\end{aligned}

Now, squaring both sides of the equation for \sin(2a) gives:

\sin^2(2a) = 4\sin^2(a)\cos^2(a) => \sin^4(2a) = 16(sin^2(a))(cos^2(a))^2

Let's represent \sin(2a)^2 and \cos(2a)^2 in terms of \sin^4(a) and \cos^4(a). This gives us:

\sin^2(2a) = 4\sin^2(a)\cos^2(a) => \sin^4(2a) = 16(sin^2(a))(cos^2(a))^2
\cos^2(2a) = \cos^4(a) - 2\sin^2(a)\cos^2(a) + \sin^4(a) => \cos^4(2a) = (\cos^4(a))^2 - 2(\sin^2(a) \cos^2(a))^2 + (\sin^4(a))^2.

Now, we can see that

\begin{aligned}
\sin^4(2a) + \cos^4(2a) &= 16(sin^2(a))(cos^2(a))^2 + (\cos^4(a))^2 - 2(\sin^2(a) \cos^2(a))^2 + (\sin^4(a))^2\\
&= 16(sin^2(a))(cos^2(a))^2 + 16(sin^2(a))^4 - 32(sin^2(a))^2(cos^2(a))^2 + 16(cos^2(a))^4.
\end{aligned}

Taking the square root yields
\sin^2(2a) + \cos^2(2a) < 4(\sin^4(a) + \cos^4(a)) => \frac{1}{2^4}\sin(2a)^4 < (\sin(a))^7 + (\cos (a))^7.

Therefore, we have shown that (\sin(a))^7 + (\cos (a))^7 < \frac{1}{2^4}\sin(2a)^4.