TrigonometryTrig. inequality: Strictly algebraic

sweatingbear

Member
Forum, do you have any idea how to solve the trigonometric inequality $$\displaystyle \cos (x) < \sin (x)$$ strictly algebraically?

The conventional(?) approach is to first solve $$\displaystyle \cos(x) = \sin(x)$$ and then draw the graphs for each function in order to find the correct interval. However, I would love to know if there is a way to do it strictly algebraically.

I attempted writing $$\displaystyle \sin (90^\circ - x) < \sin (x)$$ which yields $$\displaystyle x > 45^\circ + n \cdot 180^\circ$$ but I am unable to find an upper bound for $$\displaystyle x$$.

Klaas van Aarsen

MHB Seeker
Staff member
Forum, do you have any idea how to solve the trigonometric inequality $$\displaystyle \cos (x) < \sin (x)$$ strictly algebraically?

The conventional(?) approach is to first solve $$\displaystyle \cos(x) = \sin(x)$$ and then draw the graphs for each function in order to find the correct interval. However, I would love to know if there is a way to do it strictly algebraically.

I attempted writing $$\displaystyle \sin (90^\circ - x) < \sin (x)$$ which yields $$\displaystyle x > 45^\circ + n \cdot 180^\circ$$ but I am unable to find an upper bound for $$\displaystyle x$$.
Hi sweatingbear!

You can rewrite it as $\cos x - \sin x < 0$.

Furthermore $\cos x - \sin x = \sqrt 2 \cos(x+45^\circ)$.
See for instance the trigonometric identities on wiki.
Or you can use the cosine sum formula to verify it's true.

So $\sqrt 2 \cos(x+45^\circ) < 0$.
Know how to solve that?

MarkFL

Staff member
Perhaps try writing the inequality as:

$$\displaystyle \cos(-x)<\cos\left(270^{\circ}+x \right)$$

sweatingbear

Member
Hi sweatingbear!

You can rewrite it as $\cos x - \sin x < 0$.

Furthermore $\cos x - \sin x = \sqrt 2 \cos(x+45^\circ)$.
See for instance the trigonometric identities on wiki.
Or you can use the cosine sum formula to verify it's true.

So $\sqrt 2 \cos(x+45^\circ) < 0$.
Know how to solve that?
I have never seen that identity before, so it is rather odd in my eyes. Would it not be possible to reduce the inequality simpler one using another, more commonly seen identity? For instance the one I attempted to use?

And regarding the inequality you asked me to solve: I am not quite sure, I would really appreciate it if you could show me how you would solve that inequality. It would be very insightful to see how you do it!

Perhaps try writing the inequality as:

$$\displaystyle \cos(-x)<\cos\left(270^{\circ}+x \right)$$
Hm, which identities did you use?

Klaas van Aarsen

MHB Seeker
Staff member
I have never seen that identity before, so it is rather odd in my eyes. Would it not be possible to reduce the inequality simpler one using another, more commonly seen identity? For instance the one I attempted to use?
Well, you can also do:
$$\cos x - \sin x = \cos x - \cos(90^\circ - x)$$

And with the identity:
$$\cos p - \cos q = -2\sin\frac{p+q}2\sin\frac{p-q}2$$
It follows that:
$$\cos x - \sin x = \cos x - \cos(90^\circ - x) = -2 \sin\frac{x+(90^\circ - x)}2\sin\frac{x-(90^\circ - x)}2 = -2 \cdot \frac 12\sqrt{2} \cdot \sin(x-45^\circ)$$

And regarding the inequality you asked me to solve: I am not quite sure, I would really appreciate it if you could show me how you would solve that inequality. It would be very insightful to see how you do it!
If $\cos y < 0$, then $90^\circ < y < 270^\circ$ plus a possible multiple of $360^\circ$.

Starting from:
$$\cos(x+45^\circ)<0$$
we find:
\begin{array}{lcccl}90^\circ &<& x + 45^\circ &<& 270^\circ \\
45^\circ &<& x &<& 225^\circ
\end{array}
All plus a possible multiple of $360^\circ$.

MarkFL

Staff member
...
Hm, which identities did you use?
$$\displaystyle \cos(x)=\cos\left(0^{\circ}-(-x) \right)=\cos\left(0^{\circ} \right)\cos(-x)+\sin\left(0^{\circ} \right)\sin(-x)=\cos(-x)$$

$$\displaystyle \sin(x)=\cos\left(90^{\circ}-x \right)=\cos\left(360^{\circ}-\left(90^{\circ}-x \right) \right)=\cos\left(270^{\circ}+x \right)$$

sweatingbear

Member
Well, you can also do:
$$\cos x - \sin x = \cos x - \cos(90^\circ - x)$$

And with the identity:
$$\cos p - \cos q = -2\sin\frac{p+q}2\sin\frac{p-q}2$$
It follows that:
$$\cos x - \sin x = \cos x - \cos(90^\circ - x) = -2 \sin\frac{x+(90^\circ - x)}2\sin\frac{x-(90^\circ - x)}2 = -2 \cdot \frac 12\sqrt{2} \cdot \sin(x-45^\circ)$$

If $\cos y < 0$, then $90^\circ < y < 270^\circ$ plus a possible multiple of $360^\circ$.

Starting from:
$$\cos(x+45^\circ)<0$$
we find:
\begin{array}{lcccl}90^\circ &<& x + 45^\circ &<& 270^\circ \\
45^\circ &<& x &<& 225^\circ
\end{array}
All plus a possible multiple of $360^\circ$.
Well of course, how wonderful! Thank you so much I like Serena!

$$\displaystyle \cos(x)=\cos\left(0^{\circ}-(-x) \right)=\cos\left(0^{\circ} \right)\cos(-x)+\sin\left(0^{\circ} \right)\sin(-x)=\cos(-x)$$

$$\displaystyle \sin(x)=\cos\left(90^{\circ}-x \right)=\cos\left(360^{\circ}-\left(90^{\circ}-x \right) \right)=\cos\left(270^{\circ}+x \right)$$
Oh ok, thank you! I understand now. However, how would you solve the previously mentioned inequality?

MarkFL

Staff member
...
Oh ok, thank you! I understand now. However, how would you solve the previously mentioned inequality?
How did you solve the first version?

Prove It

Well-known member
MHB Math Helper
Forum, do you have any idea how to solve the trigonometric inequality $$\displaystyle \cos (x) < \sin (x)$$ strictly algebraically?

The conventional(?) approach is to first solve $$\displaystyle \cos(x) = \sin(x)$$ and then draw the graphs for each function in order to find the correct interval. However, I would love to know if there is a way to do it strictly algebraically.

I attempted writing $$\displaystyle \sin (90^\circ - x) < \sin (x)$$ which yields $$\displaystyle x > 45^\circ + n \cdot 180^\circ$$ but I am unable to find an upper bound for $$\displaystyle x$$.

When \displaystyle \begin{align*} -\frac{\pi}{2} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*} where \displaystyle \begin{align*} n \in \mathbf{Z} \end{align*}, we have \displaystyle \begin{align*} \cos{(x)} > 0 \end{align*}, so in this region we can say

\displaystyle \begin{align*} \sin{(x)} &> \cos{(x)} \\ \frac{\sin{(x)}}{\cos{(x)}} &> 1 \\ \tan{(x)} &> 1 \end{align*}

We know that \displaystyle \begin{align*} \tan{(x)} > 1 \end{align*} when \displaystyle \begin{align*} \frac{ \pi}{ 4} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*} and when \displaystyle \begin{align*} -\frac{3\pi}{4} + 2\pi n < x < -\frac{\pi}{2} + 2\pi n \end{align*}

So putting these together, that means we have one solution set for \displaystyle \begin{align*} \sin{(x)} > \cos{(x)} \end{align*} as \displaystyle \begin{align*} \frac{\pi}{4} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*}.

Now when \displaystyle \begin{align*} -\pi + 2\pi n < x < -\frac{\pi}{2} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*} we have \displaystyle \begin{align*} \cos{(x)} < 0 \end{align*}, so that would mean in that region we have

\displaystyle \begin{align*} \sin{(x)} &> \cos{(x)} \\ \frac{\sin{(x)}}{\cos{(x)}} &< 1 \\ \tan{(x)} &< 1 \end{align*}

We know that \displaystyle \begin{align*} \tan{(x)} < 1 \end{align*} when \displaystyle \begin{align*} -\frac{\pi}{2} + 2\pi n < x < \frac{\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} -\pi + 2\pi n < x < -\frac{3\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*}.

So putting these together, that means another possible solution set for \displaystyle \begin{align*} \sin{(x)} > \cos{(x)} \end{align*} is \displaystyle \begin{align*} -\pi + 2\pi n \leq x < -\frac{3\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*}.

Therefore the solution to \displaystyle \begin{align*} \sin{(x)} > \cos{(x)} \end{align*} is \displaystyle \begin{align*} -\pi + 2\pi n \leq x \leq -\frac{3\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{4} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*}.

sweatingbear

Member
Thanks a bunch everybody, I believe I finally understand now. Here's my rendition:

$$\displaystyle \cos (x) < \sin (x) \iff \cos (x) - \sin (x) < 0 \, .$$

The linear combination of cosine and sine, $$\displaystyle \cos (x) - \sin (x)$$, can be written as $$\displaystyle \sqrt{2} \sin (x + 135^\circ)$$. Thus we have

$$\displaystyle \sqrt{2} \sin (x + 135^\circ) < 0 \iff \sin (x + 135^\circ) < 0 \, .$$

Let $$\displaystyle t = x + 135^\circ$$. Thus $$\displaystyle \sin (t) < 0$$. If we take a look at the unit circle, we can conclude that sine is negative if and only if the argument is between $$\displaystyle 180^\circ + n \cdot 360^\circ$$ and $$\displaystyle 360^\circ + n \cdot 360^\circ$$. Therefore, we must require

$$\displaystyle 180^\circ + n \cdot 360^\circ < t < 360^\circ + n \cdot 360^\circ \, .$$

Now, substituting back $$\displaystyle t = x + 135^\circ$$ and ultimately solving for $$\displaystyle x$$ yields

$$\displaystyle 45^\circ + n \cdot 360^\circ < x < 225^\circ + n \cdot 360^\circ \, .$$

If we want to answer in the interval $$\displaystyle 0^\circ < v < 360^\circ$$, we can let $$\displaystyle p = 0$$ and finally arrive at $$\displaystyle 45^\circ < x < 225^\circ$$.

So forum, what do you think?

Klaas van Aarsen

MHB Seeker
Staff member
Looks fine to me. ;-)

You may want to support the following statement with an identity though.

The linear combination of cosine and sine, $$\displaystyle \cos (x) - \sin (x)$$, can be written as $$\displaystyle \sqrt{2} \sin (x + 135^\circ)$$.

sweatingbear

Member
Looks fine to me. ;-)

You may want to support the following statement with an identity though.
All right, great.

I used

$$\displaystyle a \sin (x) + b \cos(x) = \sqrt{a^2 + b^2} \sin \left( x + \arctan \left( \frac ba \right) \right) \, .$$

Klaas van Aarsen

MHB Seeker
Staff member
All right, great.

I used

$$\displaystyle a \sin (x) + b \cos(x) = \sqrt{a^2 + b^2} \sin \left( x + \arctan \left( \frac ba \right) \right) \, .$$
You found it!

One problem though, it's not $\arctan \left( \frac ba \right)$. It's an angle $\phi$ such that $$\displaystyle \cos \phi = \frac a {\sqrt{a^2+b^2}}$$ and $$\displaystyle \sin \phi = \frac b {\sqrt{a^2+b^2}}$$.
Or you might say $\phi = \text{atan2}(a,b)$, although not everybody is familiar with the $\text{atan2}$ function.

sweatingbear

Member
You found it!

One problem though, it's not $\arctan \left( \frac ba \right)$. It's an angle $\phi$ such that $$\displaystyle \cos \phi = \frac a {\sqrt{a^2+b^2}}$$ and $$\displaystyle \sin \phi = \frac b {\sqrt{a^2+b^2}}$$.
Or you might say $\phi = \text{atan2}(a,b)$, although not everybody is familiar with the $\text{atan2}$ function.
How is it not arctan? You can, after all, form $$\displaystyle \tan (\theta) = \frac ba$$ with your listed cosine- and sine-equations.

Klaas van Aarsen

MHB Seeker
Staff member
How is it not arctan? You can, after all, form $$\displaystyle \tan (\theta) = \frac ba$$ with your listed cosine- and sine-equations.
The problem is that $\arctan$ has a range of $(-\pi/2,\pi/2)$, while the angle you need has a range of the full $2\pi \text{ rad}$.
The function $\arctan$ fails for angles outside its range, which is slightly over half the full range.

sweatingbear

Member
The problem is that $\arctan$ has a range of $(-\pi/2,\pi/2)$, while the angle you need has a range of the full $2\pi \text{ rad}$.
The function $\arctan$ fails for angles outside its range, which is slightly over half the full range.
You are absolutely correct, thanks a lot!