Prove that there is a real value of \(x\) real such that \(\sin x+\cos x+\tan x = 0\)

[no drawn/approximated answers will be accepted]

Prove that there is a real value of \(x\) real such that \(\sin x+\cos x+\tan x = 0\)

[no drawn/approximated answers will be accepted]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewest\( \sin(x) + \cos(x) + \tan(x) = 0 \Rightarrow \sin(x) + \cos(x) = -\tan(x) \)

\( \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = \tan^2(x) \)

\( 1 + \sin(2x) = \tan^2(x) \)

\( \sin(2x) = \frac{\sin^2(x) - \cos^2(x) }{\cos^2(x)} \)

\( \sin(2x) = \frac{-\cos(2x)}{\cos^2(x)} \)

\( \tan(2x) = -\sec^2(x) \)

\( \frac{-2\tan(x)}{1-\tan^2(x)} = \left(1 + \tan^2(x) \right) \)

\( -2\tan(x) = \left(1- \tan^2(x) \right) \left(1+ \tan^2(x) \right) \)

\( -2\tan(x) = 1 - \tan^4(x) \)

Let \( y = \tan(x) \Rightarrow y^4-2y-1 = 0\)

\( y = 1 \Rightarrow y^4 - 2y - 1 = -2 < 0 \)

\( y = 2 \Rightarrow y^4 - 2y - 1 = 11 > 0 \)

\(y^4 - 2y - 1\) is a continuous function and thus by the Intermediate Value Theorem, there exists some real \(a, \; 1 < a < 2 \) such that \(y = a \Rightarrow y^4 - 2y - 1 = 0\)

This then means that since \( a \in \mathbb{R} \) then there exists some \(x \in \mathbb{R} \) such that \( \tan(x) = a\)

However, it remains to show that this result is not spurious as the second step involved squaring, and this is something I have yet to accomplish. I shall leave this here and then edit in future as appropriate. – Danny He · 2 years, 8 months ago

Log in to reply

Hmm I have a clear path to a solution which will work and give me an answer, thus solving the problem. The algebra involved is messy and complex and I think there is a much more elegant way to prove that there is a solution without proof by example. – Ali Caglayan · 2 years, 8 months ago

Log in to reply

As you can see it is not worth solving by hand. – Ali Caglayan · 2 years, 8 months ago

Here is what Mathematica gave without me chasing solutions.Log in to reply

– Aditya Raut · 2 years, 8 months ago

No need of anything at all ! You may see the solution of this problem on the website's past contests....Log in to reply

JOMO ..... – Aditya Raut · 2 years, 8 months ago

Yes, there's a good way and it will be posted ... I am a staff at the JOMO, and JOMO 5 took place over a month ago ! If you like the problems of this set (JOMO 5, and also the JOMO 6 we're gonna make now) ,,, then you should participate in theLog in to reply