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 August 22nd, 2018, 08:16 AM #1 Member   Joined: Aug 2018 From: Nigeria Posts: 43 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: 12,204 Thanks: 511 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: 43 Thanks: 2 final answer.... please
August 22nd, 2018, 09:02 AM   #4
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 August 22nd, 2018, 09:20 AM #5 Member   Joined: Aug 2018 From: Nigeria Posts: 43 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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 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.

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: 12,204 Thanks: 511 Math Focus: Calculus/ODEs skipjack...why are you editing my posts?
 August 22nd, 2018, 09:56 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,703 Thanks: 1804 Because I edited the post you quoted.
August 22nd, 2018, 09:58 AM   #9
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 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: 19,703 Thanks: 1804 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.

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