# Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$

Question:

$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\$$

$$\implies \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS}$$

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$$\implies (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\$$

$$\implies (\sin(\theta)-\cos(\theta)-1) \leq0 \because (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\$$

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$$? In this solution they have simply said: $$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$ Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

• I think it's better to use geometry for this problem. We need to determine the zone $I$ from the function $f(x)$ in the plan $Oxy$.
– NN2
Feb 7 at 20:39
• What do you mean by a plan? @NN2 Feb 15 at 7:17
• @Sid I've edited your question to avoid it having to look like an edit history . Feb 15 at 7:18
• If I understood correctly, all you want to do is prove the equivalence between the two inequalities? @Sid Feb 15 at 7:19
• nope. thats easy. the lhs part is a test qstn. i need to show where lhs>0. the rhs part is the answer key. we arent given the answer key in test.
– Sid
Feb 15 at 7:20

$$f(x)$$ has period $$2$$.

What happens in $$[0,2]$$ is repeated $$50$$ times in the same way, thus the ratio we have in this interval is the ratio of all the race.

Run points are $$1/4$$ and swim point are $$3/4$$, therefore the ratio $$swim/run$$ is $$3$$.

How did I find the points in the image below. Solve $$(\sin (2 \pi x)-3) \sin \left(\frac{\pi }{4}-\pi x\right)-\sin \left(3 \pi x+\frac{\pi }{4}\right)=2 \sqrt{2}$$ solutions are $$x=\frac12+2k;\;x=1+2k;\;k\in\mathbb{Z}$$.

In the interval $$[0,2]$$ the are $$\frac12,1$$.

Solve $$(\sin (2 \pi x)-3) \sin \left(\frac{\pi }{4}-\pi x\right)-\sin \left(3 \pi x+\frac{\pi }{4}\right)=0$$ solutions are $$x=\frac14+2k;\;\frac54+2k;\;k\in\mathbb{Z}$$

In the interval $$[0,2]$$ the are $$\frac14,\frac54$$.

The set where $$0\le f(x)\le 2\sqrt 2$$ is $$\left[\frac14,\frac12\right]\cup\left[1,\frac54\right]$$

The length is $$\frac12$$ and represents $$\frac14$$ of the interval $$[0,2]$$ while the other part is $$\frac34$$. Thus the second part is three times the first part.

$$...$$

• ? I've written exactly what you've written. The qstn was to find an intuitive way f(x) =2√2.
– Sid
Feb 8 at 3:19
• ? this is an exam. we wont have a tool to graph the function and tick the points.
– Sid
Feb 8 at 3:34
• Did I miss something ? @SomeGuy Feb 15 at 9:07
• Just thought your comment was funny, that's all @Buraian Feb 15 at 11:47

let $$t=\sin \theta -\cos \theta$$ then we get $$\sin \theta \cos \theta =\frac{1-t^2}{2}$$ indeed we have tofind $$\theta$$ which satisfies the inequality $$t(2+\frac{1-t^2}{2})\le 2$$ $$\iff f(t)=t^3-5t+4\ge 0$$ As you say this came in an exam lets presume that the factorisation is not so complicated.Suppose it has an integer root then by rational root theorem we can say that any integer root $$p$$ (say) must divide $$4$$ .$$p|4$$ so if there is an integer root it has to be one of $$\pm1,\pm2\pm3\pm4$$ We can skip $$\pm 2\pm 3\pm4$$ as $$|t|\le \sqrt{2}$$ .We are left with $$\pm 1$$ .By inspection we note that $$f(1)=0$$ hence we have found one factor i.e $$t-1$$ .The other factor can be easily found by long division ie $$t^2+t-4$$.

Now its easy to see $$t^2+t-4<0$$ for all $$|t|\le ]\sqrt{2}$$.

So $$t^3-5t+4\ge 0\implies (t-1)(t^2+t-4)\ge 0\implies t\le 1$$

Which you can be easily taken from here

• @Sid Is this fine? Feb 10 at 5:52
• yep the other guy (@steven gregory) gave more or less the same solution... i'm gonna wait and see if anyone comes up with a more intuitive way (possibly avoid cubic, if its possible, maybe by using some trigo properties to factor it somehow)... but yeah yours seems the most plausible so far.
– Sid
Feb 10 at 5:57

Hint

$$f(x)=(3-\sin (2 \pi x)) \sin \left(\pi x -\frac{\pi }{4}\right)-\sin \left(3 \pi x+\frac{\pi }{4}\right)$$ Expand and manipulate the trigonometic functions to obtain $$f(x)=\frac{7 \big[\sin (\pi x)- \cos (\pi x)\big]-\big[\sin (3 \pi x)+\cos (3 \pi x)\big]}{2\sqrt{2}}$$

• yeah that is given in the provided solution. what should we do with this result?
– Sid
Feb 8 at 7:40

## $$(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$$

Let $$x = \cos \theta$$ and $$y = \sin \theta$$. Then we get

\begin{align} (y - x)(2+xy) &\leq 2 \\ x^2 + y^2 &= 1 \end{align}

We can now argue \begin{align} -(y-x)^2 &= -1 + 2xy \\ 5 - (y-x)^2 &= 2(2 + xy) \\ \hline (y-x)(5 - (y-x)^2) &\le 4 \end{align}

Let $$z = y-x$$ and we find

\begin{align} z(5 - z^2) &\le 4 \\ z^3 - 5z + 4 \ge 0 \\ (z-1)(z^2+z-4) \ge 0 \\ \hline z \ge \dfrac{\sqrt{17}-1}{2} \\ -\frac{\sqrt{17}+1}{2} \le z \le 1 \end{align}

The first answer is superfluous. The second needs to be modified to

$$-\sqrt{2} \le z \le 1$$

$$-\sqrt{2} \le \sin \theta - \cos \theta \le 1$$

$$-\sqrt{2} \le \sqrt 2 \sin\left(x - \frac{\pi}{4} \right) \le 1$$

$$-1 \le \sin\left(x - \frac{\pi}{4} \right) \le \frac{1}{\sqrt 2}$$

$$\theta \in \left[(2n-1)\pi, \frac{\pi}{2}(4n+1) \right] \qquad \forall n \in \Bbb Z$$

• clever solution. the only issue with it is that at one point u say put "z=1." I want to completely avoid these indeterministic steps. Because if we are allowed to do that, I can simply say put sinx = 1, cosx = 0 or cosx = -1, sinx =0 in the original equation) .... and the equation is satisfied... then i can diffrentiate the given function, see where its increasing and decreasing and prove there are only 2 solutions in $x \in [-\pi, \pi]$. but the issue here is that in one step i am simply saying put sinx =1 and i want to completely avoid this (even if the solutions are obvious).
– Sid
Feb 10 at 4:40
• Where do I say to put $z=1$? Feb 10 at 4:48
• $z^3 - 5z +4 \leq 0$, i know its visible to the eye that z = 1 is a root, but even in the original equation sinx = 1 is an obvious root. i want to avoid this, if possible.
– Sid
Feb 10 at 4:53
• I'm too lazy to work it out, but I would bet that $(z-1)(z^2+z-4)$ is exactly the same information as your RHS. Feb 10 at 5:05
• yep must be, but thats what i don't understand in their solution. they factorized it without the intermediate steps. I want to find a way to factorize it by using trigonometric properties/formulae.
– Sid
Feb 10 at 5:08

Let $$A:=\sin\theta-\cos\theta,\qquad B:=\sin\theta\cos\theta$$

Then, we have $$A^2=(\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=1-2B$$

To solve $$A(2+B)-2\le 0$$

let us see if $$A(2+B)-2$$ can be factorized.

To see if $$A(2+B)-2$$ can be factorized, let us add $$k(-A^2-2B+1)$$ (which equals $$0$$) to $$A(2+B)-2$$ where $$k$$ is a non-zero real number.

\begin{align}A(2+B) -2&=2A+AB-2 \\\\&=2A+AB-2+\underbrace{k(-A^2-2B+1)}_{=0} \\\\&=k\bigg(-A^2+\frac 2kA+\frac{k-2}{k}+\frac Bk(A-2k)\bigg)\end{align}

Now we want

$$-A^2+\frac 2kA+\frac{k-2}{k}$$ to have $$A-2k$$ as a factor, so solving $$-(2k)^2+\frac 2k(2k)+\frac{k-2}{k}=0$$ gives$$k=\frac 12,\frac{-1\pm\sqrt{17}}{4}$$ So, we see that $$k=\frac 12$$ works.

Therefore, we have

\begin{align}A(2+B)-2&=A(2+B)-2+\frac 12(-A^2-2B+1) \\\\&=\frac 12\bigg(-A^2+4A-3+2B(A-1)\bigg) \\\\&=\frac 12\bigg((A-1)(-A+3)+2B(A-1)\bigg) \\\\&=\frac 12(-A+2B+3)(A-1)\end{align}

So, all we need is to solve $$\frac 12(-A+2B+3)(A-1)\le 0$$

• no no, i know how to expand the rhs and get the lhs part. i want to solve the original equation, without any extra information... say in exam condition. the only info we have is $(sinx - cosx)(2+sinxcosx) \leq 2$... my question is not to prove if LHS = RHS. we are only given LHS, and we need to find theta.
– Sid
Feb 10 at 5:15
• @Sid : Thanks. I completely rewrote my answer. Feb 10 at 5:37
• yep seems legit, but the step of adding that random factor seems so unintuitive right?... and by the way, you can solve the given equation in any way u wish... u dont necessarily need to go the same route as given in the answer key (i.e. arrive at the RHS part)... u can arrive at a different expression as well and solve the qstn. I'm just looking for an intuitive solution (if it exists)
– Sid
Feb 10 at 5:43

We will use

$$(\sin\theta- \cos\theta)^2 = \sin^2\theta - 2\sin\theta \cos\theta + \cos^2\theta = 1 - 2\sin\theta \cos\theta$$ a few times.

$$(\sin \theta -\cos \theta)(2+\sin \theta \cos \theta)- 2 \le 0$$

$$2[(\sin \theta -\cos \theta)(2+\sin \theta \cos \theta)- 2] \le 0$$

$$(\sin \theta -\cos \theta)(4+2\sin \theta \cos \theta)- 4 \le 0$$

$$(\sin \theta -\cos \theta)(5-(1-2\sin \theta \cos \theta))- 4 \le 0$$

$$(\sin \theta -\cos \theta)(5-(\sin\theta - \cos\theta)^2)- 4 \le 0$$

$$-(\sin\theta - \cos\theta)^3 + 5(\sin\theta - \cos\theta)-4 \le 0$$

$$(\sin\theta - \cos\theta - 1) (-(\sin\theta - \cos\theta)^2 - (\sin\theta - \cos\theta) + 4) \le 0$$

$$(\sin\theta - \cos\theta - 1) (-1 + 2\sin\theta\cos\theta - \sin\theta + \cos\theta) + 4) \le 0$$

$$(\sin\theta - \cos\theta - 1) (\cos\theta - \sin\theta + 2\sin\theta\cos\theta + 3) \le 0$$

• in the third last line, how did u factor out sinx - cosx -1...
– Sid
Feb 10 at 10:53
• Let $x=\sin \theta - \cos\theta$ then I can compute $\dfrac{-x^3+5x-4}{x-1}$ using synthetic division. I know $x=1$ is a root because $-x^3+5x-4 = 0$ when $x=1$. Feb 10 at 11:02
• Yeah but how do u know 1 is a root. If we are allowed to do that, I can say put sinx=1 in the original equation... That's also an obvious root. The issue here is that we are guessing
– Sid
Feb 10 at 11:04

Imagine radius vector rotation in the four quadrants

At $$\theta$$ (0,180) sine vanishes

At $$\theta$$ (90,270 ) cos vanishes

So the four zones are determined and demarcated as four the quadrants conveniently.

$$F(\theta)=(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) - 2$$

For extremum values only center values of each quad $$F(\theta)$$ needs to be evaluated/examined that too only for its sign.

We need only to evaluate at odd central values

$$(2k-1)\dfrac{\pi}{4}$$

$$0 (...) -2 \le 0 \text{ okay}$$

$$(2/\sqrt2)(2-\frac12) >0 \text{ not okay}$$

$$(2\sqrt2)(2-1/\sqrt2) \le 0 \text { okay}$$

$$(2 \sqrt2)(....) \le 0 \text { okay}$$

So the correct sign occurs in 1st, 2nd, 4th quadrants only.

$$(\pi/2<\theta<3 \pi/2 )$$ interval is inadmissible.

No need for a polar graph to be drawn... even if drawn for confirmation it would not be single valued radius in the second quadrant anyways..

I have re-written this answer because, with the help of WolframAlpha here, I was surprised to discover

$$0\le \big(3-\sin(2\pi x)\big)\sin(\pi x - \frac{\pi}{4})- \sin(3\pi x +\frac{\pi}{4})\le 2\sqrt{2}\\ \implies (2 n - 1)

What the first "implies" means for $$\space 0\le x\le 100\space$$ is that there are discreet ranges of values that make this true as in these samples for $$1\le n\le 6$$

$$(1\le x \le \frac{5}{4}=1.25)\quad (3\le x \le \frac{13}{4}=3.25)\quad (5\le x \le \frac{21}{4}=5.25)\quad \\ (7\le x \le \frac{29}{4}=7.25)\quad (9\le x \le \frac{37}{4}=9.25)\quad (11\le x \le \frac{45}{4}=11.25)\quad$$

What the second "implies" means is that the following ranges apply for $$\space x$$ for $$1\le n\le 6$$.

$$\frac{9}{4}=2.25\le x \le \frac{5}{2}=2.5\qquad \frac{17}{4}=4.25\le x \le \frac{9}{2}=4.5\\ \frac{25}{4}=6.25\le x \le \frac{13}{2}=6.5\qquad \frac{33}{4}=8.25\le x \le \frac{17}{2}=8.5\\ \frac{41}{4}=10.25\le x \le \frac{21}{2}=10.5\qquad \frac{49}{4}=12.25\le x \le \frac{25}{2}=12.5$$

We can see that each value of $$n$$ generates $$\space x\space$$ with a total "coverage of $$\space 0.5\qquad$$ Note that $$\quad 1\le n \le 24\implies 1\le x \le 97\quad$$ so there are $$24\times0.5=12$$ "full-numbers-worth" of $$0\le x \le 100$$ that satisfy $$I.\quad$$ However, the first "implies" above permits $$n=25\implies 99\le x \le 99.25\space$$ and the second "implies" peermits $$n=0.0\implies 0.25\le x \le 0.5\quad$$. There are a total of $$100$$ intervals of $$0.25$$ each so the members run $$\frac14$$ of the time.

This means that there members swim $$\frac34$$ of the time and the ratio of $$\frac{swim}{run}$$ is $$\quad 3:1$$

• Comments are not for extended discussion; this conversation has been moved to chat. Feb 10 at 23:35