MHB Solve a trigonometric equation

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The discussion centers on solving the trigonometric equation $\tan 4y=\frac{\cos y-\sin y}{\cos y+\sin y}$ for $y$ in the interval $0<y<\frac{\pi}{4}$. A participant demonstrates the solution by applying the tangent subtraction formula, specifically $\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$, with $a=\frac{\pi}{4}$ and $b=y$. The conversation highlights the use of this formula rather than rote memorization to derive the solution. Overall, the focus is on the application of trigonometric identities to solve the equation. The discussion effectively illustrates problem-solving techniques in trigonometry.
anemone
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Let $y$ be in radians and $0<y<\dfrac{\pi}{4}$.

Solve for $y $ if $\tan 4y=\dfrac{\cos y-\sin y}{\cos y +\sin y}$.
 
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anemone said:
Let $y$ be in radians and $0<y<\dfrac{\pi}{4}$.

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

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
$$y=\frac{\pi}{20}$$
 
Pranav said:
$$\frac{1-\tan y}{1+\tan y}=\tan\left(\frac{\pi}{4}-y\right)$$

How did you get that?
It's not something you had to learn by heart did you? :rolleyes:
 
I like Serena said:
It's not something you had to learn by heart did you? :rolleyes:

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 $b=y$. :)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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