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 August 22nd, 2018, 08:16 AM #1 Member   Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 Trigonometry Given tan(x-y)=-2/3 and tanx=-1/2, find the value of a. tan y b. (x+y) for 0
 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 well-known identity (angle-difference for tangent) tells us: $\displaystyle \tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$ 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
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Quote:
 Originally Posted by Harmeed final answer.... please
I'm trying to help you find the answer, not just do your homework for you. That doesn't help you learn. August 22nd, 2018, 09:20 AM #5 Member   Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 But I obtained -3-3y/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
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Quote:
 Originally Posted by Harmeed But I obtained -3-3y/2, but my textbook says 1/8. I need full workings.
Without seeing your actual work, I have no idea what you did wrong, and the expression you supplied makes no sense either, since it is not an equation.

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 $\displaystyle \frac{2}{2}$:

$\displaystyle \frac{1+2u}{2-u}=\frac{2}{3}$

Multiply through by $\displaystyle 3(2-u)$:

$\displaystyle 3(1+2u)=2(2-u)$

Distribute:

$\displaystyle 3+6u=4-2u$

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
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Quote:
 Originally Posted by skipjack Because I had edited the post you quoted.
Okay...I notice many of my posts get edited by you, and I've been curious about it. Today, I decided to inquire. 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 right-hand side by using the tangent of angle-difference 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. Tags trigonometry Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post John Marsh Trigonometry 3 September 6th, 2013 02:27 AM kiyoshi7 Trigonometry 3 August 29th, 2013 08:30 AM johnny Algebra 6 March 7th, 2011 01:06 PM johnny Algebra 7 March 5th, 2011 08:39 AM willab Algebra 2 October 11th, 2010 01:11 PM

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