Solve a trigonometric equation

[anemone]

Let $y$ be in radians and $0<y<\dfrac{\pi}{4}$.

Solve for $y $ if $\tan 4y=\dfrac{\cos y-\sin y}{\cos y +\sin y}$.

 

[Saitama]

Let $y$ be in radians and $0<y<\dfrac{\pi}{4}$.

Solve for $y $ if $\tan 4y=\dfrac{\cos y-\sin y}{\cos y +\sin y}$.

Spoiler

We can rewrite the RHS as:
$$\tan 4y=\frac{1-\tan y}{1+\tan y}=\tan\left(\frac{\pi}{4}-y\right)$$
$$\Rightarrow 4y=n\pi+\frac{\pi}{4}-y$$
Only n=0 gives a solution in the specified range, hence
$$y=\frac{\pi}{20}$$

 

[I like Serena]

$$\frac{1-\tan y}{1+\tan y}=\tan\left(\frac{\pi}{4}-y\right)$$

How did you get that?
It's not something you had to learn by heart did you?

 

[Saitama]

It's not something you had to learn by heart did you?

Nope. :D

I used the following formula:
$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$
with $a=\pi/4$ and $b=y$. :)

 

[CellCoree]

To solve this trigonometric equation, we can start by simplifying the right side of the equation using trigonometric identities. We can rewrite the numerator as $\cos y - \sin y = \cos y \left(1 - \tan y\right)$ and the denominator as $\cos y + \sin y = \cos y \left(1 + \tan y\right)$. Substituting these into the original equation, we get:

$\tan 4y = \dfrac{\cos y \left(1 - \tan y\right)}{\cos y \left(1 + \tan y\right)}$

Next, we can simplify the equation by dividing both sides by $\cos y$:

$\tan 4y = \dfrac{1 - \tan y}{1 + \tan y}$

We can then use the double angle formula for tangent, $\tan 2\theta = \dfrac{2\tan \theta}{1-\tan^2 \theta}$, to rewrite the equation as:

$\tan 4y = \tan 2y$

Since $0<y<\dfrac{\pi}{4}$, we know that $0<2y<\dfrac{\pi}{2}$, which means that the tangent function has a one-to-one correspondence in this interval. Therefore, we can set the arguments of the tangent functions equal to each other:

$4y = 2y$

Solving for $y$, we get $y=0$. However, this solution does not satisfy the given interval of $0<y<\dfrac{\pi}{4}$. Therefore, there is no solution for $y$ that satisfies the given equation within the given interval. This can also be seen by graphing the two sides of the equation, which do not intersect within the given interval.