Find $\tan(2x)$ Given $\cos(x-y)$ and $\sin(x+y)$

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In summary, the formula for finding the tangent of 2x is tan(2x) = (2tan(x))/(1-tan^2(x)), and this can be used to find the value of tan(2x) when given values for cos(x-y) and sin(x+y). The relationship between tan(2x), cos(x-y), and sin(x+y) can be understood using the double angle identities for tangent, cosine, and sine. To simplify the expression tan(2x), one can use the formula and substitute the given values, then simplify using algebraic manipulations. Additionally, there are alternative methods for finding the value of tan(2x) such as using the half angle identities for tangent or the Py
  • #1
kaliprasad
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If $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$ and x,y are between 0 and $\frac{\pi}{4}$
then find $\tan (2x)$
 
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  • #2
My solution:

\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)=\frac{3}{5}\rightarrow \tan(x-y)=\frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}=\frac{56}{33}\]
 
  • #3
lfdahl said:
My solution:

\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)=\frac{3}{5}\rightarrow \tan(x-y)=\frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}=\frac{56}{33}\]

Above solution is correct but one solution is missing
 
  • #4
kaliprasad said:
Above solution is correct but one solution is missing

I´m sorry for the missing answer. Below is (hopefully) the second solution included:

\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)= \pm \frac{3}{5}\rightarrow \tan(x-y)=\pm \frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\pm \frac{3}{4}+\frac{5}{12}}{1 \mp \frac{3}{4}\cdot \frac{5}{12}}= \left\{\begin{matrix} \: \: \: \frac{56}{33}\\ \\ -\frac{16}{63} \end{matrix}\right.\]
 

FAQ: Find $\tan(2x)$ Given $\cos(x-y)$ and $\sin(x+y)$

What is the formula for finding the tangent of 2x given cosine of x-y and sine of x+y?

The formula for finding the tangent of 2x is tan(2x) = (2tan(x))/(1-tan^2(x)), where tan(x) = sin(x)/cos(x).

How do you find the value of tan(2x) when given the values of cos(x-y) and sin(x+y)?

To find the value of tan(2x), first use the formula tan(2x) = (2tan(x))/(1-tan^2(x)). Then, substitute the values of cos(x-y) and sin(x+y) into the formula for tan(x).

Can you explain the relationship between tan(2x), cos(x-y), and sin(x+y)?

The relationship between tan(2x), cos(x-y), and sin(x+y) can be understood by using the double angle identities for tangent, cosine, and sine. Specifically, tan(2x) = (2tan(x))/(1-tan^2(x)), cos(x-y) = cos(x)cos(y) + sin(x)sin(y), and sin(x+y) = sin(x)cos(y) + cos(x)sin(y).

How do you simplify the expression tan(2x) when given values for cos(x-y) and sin(x+y)?

To simplify the expression tan(2x), first use the formula tan(2x) = (2tan(x))/(1-tan^2(x)). Then, substitute the values of cos(x-y) and sin(x+y) into the formula for tan(x). Finally, simplify the resulting expression using algebraic manipulations.

Are there any alternative methods for finding the value of tan(2x) when given cosine of x-y and sine of x+y?

Yes, there are some alternative methods for finding the value of tan(2x) when given cosine of x-y and sine of x+y. One method is to use the half angle identities for tangent, which states that tan(2x) = (sin(2x))/(cos(2x)) = (2sin(x)cos(x))/(cos^2(x)-sin^2(x)). Another method is to use the Pythagorean identity for tangent, which states that tan^2(x) = sec^2(x) - 1.

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