Homework Help: Trigonometry

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

Find the exact value of tan(a-b)

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 11:54pm
Both 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 12:16am
I put that in but it said the answer was
36-25Бу2/23
any idea how that works?
Trigonometry - Reiny, Tuesday, November 9, 2010 at 12:40am
That is exactly what my answer works out to if you rationalize it.
I had suggested to do that.

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