Finding the exact value?

Question

My math teacher gave us a lovely home assignment to finish on our own. I'm absolutely stuck on a couple. If you could point me in the right direction on one, I could probably get the others.
Could you help me with this, please?

Given that tan a = - 4/3, a lies in the quadrant II and cos b =2/3, b lies in quadrant I. Find the exact value of sin(a+b)

I know now that
cos(a) = -3/5
sin(a) = 4/5
cos(b) = 2/3
sin(b) = sqrt(5)/3

What if I'm trying to solve for tan(a+b)?

Thanks for all the help Puggy and Wal C! You guys are great.
What if I want to solve for tan(a+b)?

Answer 1

Given that tan a = - 4/3, a lies in the quadrant II, and cos b =2/3, b lies in quadrant I. Find the exact value of cos(a+b)

tan a = -4/3

tan²a = 16/9

tan²a + 1 = 25/9 = sec²a

a lies in quadrant II data so seca,cosa < 0 and sina > 0

So sec a = -5/3

Thus cosa = -3/5 (as sin²a + cos²a = 1
→ sina = ±√(1 - cos²a)

and sina = +√(1 - 9/25) (a is in quadrant II)
= √(16/25)
= 4/5

cosb = 2/3

b is in quadrant l so sinb>0

sinb = +√(1 - 4/9) (since sin²b + cos²b = 1)
= √(5/9)
= √5/3

Now cos(a + b) = cosa cosb - sina sinb
= -3/5 * 2/3 - 4/5 * √5/3

= -(6 + 4√5)/15

..... 6 + 4√5
= – ―――― (.... are just spacers)
...... 15

and sin(a + b) = sina cosb + cosa sinb
= 4/5 * 2/3 + -3/5 * √5/3

= (8 - 3√5)/15

__________________________ _________________________


Now should you also need tan(a + b) (How was that for anticipation hahahahahahaha!!!!!)

Tan(a + b) = sin(a + b)/cos(a + b)

= (8 - 3√5)/15 ÷ -(6 + 4√5)/15

= - (3√5 - 8)/(6 + 4√5) * (6 - 4√5)/(6 - 4√5) (... See below)

= (60 + 48 + 18√5 + 32√5)/(36 - 80)

= (108 - 50√5)/(-44)

= (25√5 - 54)/22
__________________________ _________________________

(.... from above)

- (3√5 - 8)(6 - 4√5)
---------------------------- =
(6 + 4√5)(6 - 4√5)

60 + 48 - 50√5
--------------------- =
36 - 80

108 - 50√5
---------------- =
-44

25√5 - 54
---------------
22

Answer 2

The only reason why you were able to solve your last question (which was solving for cos(a + b)) was because of the identity

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

In this case there's a formula for sin(a + b) as well [refer to your textbook]:

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

Plugging in your proposed values,
sin(a + b) = [4/5] [2/3] + [sqrt(5)/3][-3/5]
sin(a + b) = [8/15] - sqrt(5)/5

To simplify it all under one fraction,

sin(a + b) = [8 - 3sqrt(5)]/5

Answer 3

sin(a+b) = sin(a)cos(b) + sin(b)cos(a)
but we know that
sin(b) = Sqrt(1-cos(b)^2)
and we also know that
sin(a)/cos(a) = tan(a)
from this, we can derive
sin(a) = tan(a)/Sqrt(1+tan(a)^2)
and
cos(a) = 1/Sqrt(1+tan(a)^2)
So, the final equation in terms of tan(a) and cos(b) is:

(tan(a)/Sqrt(1+tan(a)^2)) (cos(b))
+ (Sqrt(1-cos(b)^2) (1/Sqrt(1+tan(a)^2)

((-4/3)/Sqrt(1+16/9)) (2/3) + (Sqrt(1-4/9) (1/Sqrt(1+16/9)) =

-8/15 + Sqrt(3/5)

... man, which one of us has got this right?

Answer 4

the best cost of tan210 is (a million/2) / (the sq. root of three divided by using 2 2) sorry idk the thank you to do the foundation image on my workstation. hah. i purely finished up a trig unit in my class like 3 weeks in the past so this could be good.

Answer 5

*use the identities.
sin(a+b) =sina*cosb+ cosa*sinb

Answer 6

so sin(a+b) = sin(a) cos(b) + cos(a) sin(b);