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Pre-calculus/trig

 Apr 1, 2019
 #1
avatar+19326 
+2

Here is a webpage .......look at   Hal Angle formulas and see what you find.

http://home.windstream.net/okrebs/page103.html

 

Remember   Sin and COs are both negative in Quadrant III

sin = -4/5                                     cos = - 3/5     (because cos^2 = 1- sin^2)

 

Then go from there ......

 Apr 1, 2019
 #2
avatar+315 
0

tan(x/2)=(8/5)/(-4/5)

tan(x/2)=(8/5)(-5/4)

tan(x/2)=-2

 

Did I do the right evaluation? 

 

So, wouldn't that make the answer -2?

Cupcake  Apr 1, 2019
edited by Cupcake  Apr 1, 2019
 #3
avatar+19326 
+1

I'll use the first evalution to see if I get the same answer as you did:

Tan x/2 = +-  sqrt (   (1-cosx) /(1+cos x) )

              =+-sqrt (  (1- -3/5)/(1+ -3/5)  )   = + -  sqrt( 8/8  /  2/5)  = +-sqrt  (4)  =+2 or  -2       We know it is the negative one because it is in Q II

 

So YES!   The answer is  -2  !!!

 Apr 1, 2019
edited by ElectricPavlov  Apr 1, 2019
 #4
avatar+315 
0

So, I'm right?

Cupcake  Apr 1, 2019
 #5
avatar+315 
0

Thank you!! smiley

Cupcake  Apr 1, 2019

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