Posted by **Sophie** on Tuesday, June 9, 2009 at 1:36pm.

If α and β are two angles in Quadrant II such that tan α= -1/2 and tan β = -2/3, find cos(α+β)

Work:

cos(α+β) = [ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)]

cos(α+β) = [ 1 - (-1/2)(-2/3) ] / [ 1 + (-1/2)(-2/3)]

cos(α+β) = [ 1 - 1/3 ] / [ 1 + 1/3 ]

cos(α+β) = [ 2/3 ] / [ 4/3 ]

cos(α+β) = 1/2

is this right?

- Math -
**MathMate**, Tuesday, June 9, 2009 at 4:47pm
Unfortunately it doesn't look right, ad... from the very start.

Could you rechceck your source from where you got the expression?

[ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)]

represents, according to me,

cos(α+β)/cos(α-β).

All is not lost, though.

From

1/cos^{2}(x)

=sec^{2}(x)

=1+tan^{2}(x)

tan(A+B)=(tan(A)+tan(B))/(1-tan(A)tan(B))

You have ample ammunition to get the required answer.

Post your results for a check-up if you wish, or you can substitute arbitrary numbers for tan(α) and tan(β) to check by yourself.

- Math -
**Sophie**, Tuesday, June 9, 2009 at 8:18pm
ohh okay thanks!

- Math -
**Sophie**, Tuesday, June 9, 2009 at 8:29pm
tan^-1(1/2)=x=.463648

tan^-1(2/3)=y.588003

x+y+pi/2 = -z=2.62245

z=-2.62245

cos (z) = -.868243

and

tan a= -1/2

tan b= -2/3

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

so tan (a+b)= -7/4

sec^2 x-tan^2 x=1

so

sec (a+b)=√(1+49/16)

=√(65/16)

(a+b)=1/sec (a+b)

so cos(a+b) is positive

so cos(a+b)=.48˜.5

???

- Math -
**MathMate**, Wednesday, June 10, 2009 at 12:14am
tan^-1(1/2)=x=.463648

**α=tan**_{-1}(-1/2)=π-x=2.67795

tan^-1(2/3)=y.588003

**β=tan**_{-1}(-2/3)=π-y=2.55359

**α+β = 2.67795+2.55359 = 5.23154 rad. (serves as a check)
**

cos(α+β) = 0.49614 (as a check)

and

tan a= -1/2

tan b= -2/3

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

so tan (a+b)= -7/4

sec^2 x-tan^2 x=1

so

sec (a+b)=√(1+49/16)

=√(65/16)

** Therefore
**

cos(α+β)

=√(16/65)

=4/√65

since α+β is in the 4th quadrant, cos(α+β)>0

Therefore

cos(α+β) = +4/√65

- Math -
**Sophie**, Wednesday, June 10, 2009 at 2:28pm
oh! thanks for your time!

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