August 22nd, 2018, 08:16 AM  #1 
Member Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2  Trigonometry
Given tan(xy)=2/3 and tanx=1/2, find the value of a. tan y b. (x+y) for 0<x+y<360 please. I need all possible solutions. Last edited by skipjack; August 22nd, 2018 at 10:20 AM. 
August 22nd, 2018, 08:24 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
A wellknown identity (angledifference for tangent) tells us: $\displaystyle \tan(xy)=\frac{\tan(x)\tan(y)}{1+\tan(x)\tan(y)}$ So, equate this to the given value, and substitute for $\displaystyle \tan(x)$ to get: $\displaystyle \frac{\dfrac{1}{2}\tan(y)}{1\dfrac{1}{2}\tan(y)}=\frac{2}{3}$ Now, solve for $\displaystyle \tan(y)$. 
August 22nd, 2018, 08:55 AM  #3 
Member Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 
final answer.... please

August 22nd, 2018, 09:02 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  
August 22nd, 2018, 09:20 AM  #5 
Member Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 
But I obtained 33y/2, but my textbook says 1/8. I need full workings.
Last edited by skipjack; August 22nd, 2018 at 09:51 AM. 
August 22nd, 2018, 09:42 AM  #6  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Quote:
Let's let $\displaystyle u=\tan(y)$ and write: $\displaystyle \frac{\dfrac{1}{2}u}{1\dfrac{1}{2}u}=\frac{2}{3}$ Multiply both sides by $\displaystyle 1$: $\displaystyle \frac{\dfrac{1}{2}+u}{1\dfrac{1}{2}u}=\frac{2}{3}$ Multiply the LHS by $\displaystyle \frac{2}{2}$: $\displaystyle \frac{1+2u}{2u}=\frac{2}{3}$ Multiply through by $\displaystyle 3(2u)$: $\displaystyle 3(1+2u)=2(2u)$ Distribute: $\displaystyle 3+6u=42u$ Collect like terms: $\displaystyle 8u=1$ Divide through by 8 to obtain: $\displaystyle u=\tan(y)=\frac{1}{8}$ To do part b), use the angle sum identity for tangent, and use the now known values of $\displaystyle \tan(x)$ and $\displaystyle \tan(y)$ to get a numerical value of $\displaystyle \tan(x+y)$, from which you can deduce the possible values of the angle $\displaystyle x+y$ on the given interval. This time, please show each step of your work. Last edited by skipjack; August 22nd, 2018 at 09:51 AM.  
August 22nd, 2018, 09:54 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
skipjack...why are you editing my posts?

August 22nd, 2018, 09:56 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2203 
Because I edited the post you quoted.

August 22nd, 2018, 09:58 AM  #9 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  
August 22nd, 2018, 10:07 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2203 
As tan(y) = tan(x  (x  y)), expand the righthand side by using the tangent of angledifference formula once, then substitute the values given for tan(x  y) and tan(x) and evaluate the result. Can you also post your work? After tan(y) has been found, expand tan(x + y) in terms of tan(x) and tan(y), then substitute the values of tan(x) and tan(y). Use this result to find a possible value (in degrees, to an appropriate number of decimal places) of x + y. To find values in the specified range, add or subtract integer multiples of 180 degrees. 

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