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September 18, 2014

September 18, 2014

Posted by **Patti** on Saturday, February 19, 2011 at 11:21pm.

- Trig -
**MathMate**, Sunday, February 20, 2011 at 8:55ams is in Q1 (sin>0, cos>0)

so

cos(s)=1/3,

sin(s)=+√(1-(1/3)²)

=√(8/9)

tan(s)=√(8/9) / (1/3)

=√8

t is in Q4 (sin<0, cos>0)

sin(t)=-1/2

cos(t)=+√(1-(1/2)²)

=√(3/4)

tan(t)=(-1/2)/√(3/4)

=-1/√3

[rationalize by multiplying numerator and denominator by √3]

=-(√3)/3

Use the identity

tan(s+t)=(tan(s)+tan(t))/(1-tan(s)tan(t)) to calculate tan(s+t)

I get about 0.85.

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