The utility of trigonometrical substitutions

Today we will discuss the power of trigonometrical substitutions.

Let us take the expression \frac{\sum_{k=1}^{2499} \sqrt{10+\sqrt{50+\sqrt{k}}}}{\sum_{k=1}^{2499} \sqrt{10-\sqrt{50+\sqrt{k}}}}

This is a math competition problem. One solution proceeds this way: let p_k=\sqrt{50+\sqrt{k}}, q_k=\sqrt{50-\sqrt{k}}. Then as p_k^2+q_k^2=10^2, we can write p_k=10\cos x_k and q_k=10\sin x_k.

This is an elementary fact. But what is the reason for doing so?

Now we have a_k=\sqrt{10+\sqrt{50+\sqrt{k}}}=\sqrt{10+10\cos x_k}=\sqrt{20}\cos \frac{x_k}{2}. Similarly, b_k=\sqrt{10-\sqrt{50+\sqrt{k}}}=\sqrt{10-10\cos x_k}=\sqrt{20}\sin \frac{x_k}{2}. The rest of the solution can be seen here. It mainly uses identities of the form 2\sin A\cos B=(\sin A+\cos B)^2 to remove the root sign.

What if we did not use trigonometric substitutions? What is the utility of this method?

We will refer to this solution, and try to determine whether we’d have been able to solve the problem, using the same steps, but not using trigonometrical substitutions.


=\sqrt{10+10(\frac{\sqrt{50+\sqrt{k}}}{10\sqrt{2}}+\frac{\sqrt{50-\sqrt{k}}}{10\sqrt{2}})}=\sqrt{10+10\times 2\times\frac{\frac{1}{2}(\frac{\sqrt{50+\sqrt{k}}}{10\sqrt{2}}+\frac{\sqrt{50-\sqrt{k}}}{10\sqrt{2}})}{\sqrt{\frac{50+\sqrt{k}}{20\sqrt{2}}-\frac{50-\sqrt{k}}{20\sqrt{2}}+\frac{1}{2}}}\times \sqrt{\frac{50+\sqrt{k}}{20\sqrt{2}}-\frac{50-\sqrt{k}}{20\sqrt{2}}+\frac{1}{2}}}

As one might see here, our main aim is to remove the square root radicals, and forming squares becomes much easier when you have trigonometrical expressions. Every trigonometrical expression has a counterpart in a complex algebraic expression. It is only out of sheer habit that we’re more comfortable with trigonometrical expressions and their properties.

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Graduate student

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