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January 26, 2015

January 26, 2015

Posted by **Dee** on Monday, November 8, 2010 at 11:40pm.

sin a = 4/5, -3pi/2<a<-pi;

tan b = -sqrt2, pi/2<b<pi

identity used is:

tan(a-b)=(tan a-tan b)/1+tan a tan b

simplify answer using radicals.

(a is alpha, b is beta)

- Trigonometry -
**Reiny**, Monday, November 8, 2010 at 11:54pmBoth the sine and cosine curves are the same for

-3pi/2<a<-pi as they are for π/2<a<π (2nd quad)

so if sina = 4/5, then cosa = -3/5

and tana = -4/3

then tan(a-b)

= (tana - tanb)/(1 + tanatanb)

= ((-4/3) - (-√2))/( 1 + (-4/3)(-√2))

= (√2 - 4/3)/(1 + 4√2/3)

= (3√2 - 4)/(3 + 4√2)

I don't know if you have to rationalize that, if you do carefully multiply top and bottom by (3 - 4√2)

- Trigonometry -
**Dee**, Tuesday, November 9, 2010 at 12:16amI put that in but it said the answer was

36-252/23

any idea how that works?

- Trigonometry -
**Reiny**, Tuesday, November 9, 2010 at 12:40amThat is exactly what my answer works out to if you rationalize it.

I had suggested to do that.

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