Trigonometry! #100

Geometry Level 3

$a_{1}+a_{2}\cos(2x)+a_{3}\sin^{2}(x)=0$

How many ordered triplets $(a_1, a_2, a_3 )$ satisfy the above equation for all $x$ ?

This problem is part of the set Trigonometry .

None of these Three Infinite One Zero

1 solution

Brian Charlesworth
Feb 9, 2015

Let $(a_{1}, a_{2}, a_{3}) = (-k, k, 2k)$ for any real $k$ . We then have that

$a_{1} + a_{2}\cos(2x) + a_{3}\sin^{2}(x) =$

$-k + k(\cos^{2}(x) - \sin^{2}(x)) + 2k\sin^{2}(x) =$

$-k + k(\cos^{2}(x) + \sin^{2}(x)) = -k + k = 0.$

Since this holds true for any real $k$ , we can conclude that there are an infinite number of possible triplets.

How did you claim the very first statement?

Raven Herd - 6 years, 3 months ago

I noticed after working with the expression on the LHS of the equation briefly that if I let $(a_{1}, a_{2}, a_{3}) = (-k.k.2k)$ then the expression always came out $0$ . This gave me an infinite number of triplets so I didn't have to go any further, although I believe that any triplet that satisfies the equation for all $x$ will probably have to be of this form, since it appears to be the only way of eliminating the variability of the trigonometric terms.

Brian Charlesworth - 6 years, 3 months ago

Thank you sir,that was really insightful.

Raven Herd - 6 years, 3 months ago

But we can also deduce that since expression is true for all x , keep x =0, hence given expression becomes a {1} + a {2} + a_{3} = 0 Which, seemingly, has infinitely many solutions :)

Ravi Mistry - 6 years, 3 months ago

We have to find triplets $(a_{1}, a_{2}, a_{3})$ so that the statement is true for all $x$ , not just $x = 0$ .

For $x = 0$ the equation would actually become $a_{1} + a_{2} = 0$ , which does have infinitely many solutions, but this would only be the case for $x = 0$ , any hence would not necessarily hold true when we allow $x$ to vary over all real numbers. My solution finds triplets that make the equation true for all real $x$ as required. The key is to find triplets that eliminate the effect of the variability of $x$ . Hope this makes sense. :)

Brian Charlesworth - 6 years, 3 months ago

Trigonometry! #100

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