×

# JOMO 5, Long 2

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]

3 years ago

Sort by:

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

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

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

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

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

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

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

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

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

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. · 3 years ago

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. · 3 years ago

Here is what Mathematica gave without me chasing solutions. As you can see it is not worth solving by hand. · 3 years ago

No need of anything at all ! You may see the solution of this problem on the website's past contests.... · 3 years ago