+0

# Pre-calculus/trig

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5 Pre-calculus/trig

Apr 1, 2019

#1
+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
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
+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
0

So, I'm right?

Cupcake  Apr 1, 2019
#5
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Thank you!! Cupcake  Apr 1, 2019