Help with 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 cos(a+b)

Thanks for all the help Puggy, Wal C, and M.Abuhelwa! You guys are great.

What if I want to solve for tan(a+b)?

Answer 1

Since you're given tan(a) = -4/3, and tan = opp/adj, make a right angle triangle with angle a and opposite side equal to 4 and adjacent side equal to 3. By the Pythagorean theorem, the hypotenuse should be equal to sqrt (3^2 + 4^2) = sqrt (25) = 5.
For that reason, it follows that sin(a) = opp/hyp = 4/5 (it is positive because in quadrant 2, sine is positive).

Similarly, cos(a) = adj/hyp = -3/5. (It is negative since "a" is given to be in quadrant 2, where cosine is negative).

Since cos(b) = 2/3, and cos = adj/hyp, make a right angle triangle with adj = 2 and hyp = 3. It then follows that
opp = sqrt(3^2 - 2^2) = sqrt(9 - 4) = sqrt(5).

Therefore, sin(b) = opp/hyp = sqrt(5)/3

By the cos addition identity,

cos(a + b) = cosacosb - sinasinb

And since we know that
cos(a) = -3/5
sin(a) = 4/5
cos(b) = 2/3
sin(b) = sqrt(5)/3

cos(a + b) = (-3/5)(2/3) - (4/5) (sqrt(5)/3)
cos(a + b) = (-2/5) - (4sqrt(5)/15)
cos(a + b) = (-6 - 4sqrt(5)]/15

Answer 2

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 secb,cosb < 0 and sinb > 0

So sec a = -5/3

Thus cosa = -3/5

and sina = +√1 - (9/25) (as sin²a + cos²a = 1)
= √(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

Answer 3

To solve this, you first need to determine sine and cosine of both angles a and b.

Angle a is in the second quadrant so sine is positive and cosine is negative.

tan a = sin a / cos a = -4/3 = (4/x)/(-3/x)

Solving for x:

sin² a + cos² a = 16/x² + 9/x² = 25/x² = 1
x² = 25
x = 5

cos a = -3/5
sin a = 4/5

Angle b is in the first quadrant so both sine and cosine are positive.

cos b = 2/3

sin² b + cos² b = 1
sin² b = 1 - cos² b = 1 - 2/3 = 1/3
sin b = 1/√3

Now we will calculate cos(a + b)

cos(a + b) = (cos a)(cos b) - (sin a)(sin b)
= (-3/5)(2/3) - (4/5)(1/√3)
= -2/5 - (4/5)(1/√3)
= -2/5 - 4/(5√3)
= -6/15 -4√3/15 = -(6 + 4√3)/15

Answer 4

use pethagorian rule to get
opposit(O) ,adjacent(A) , and hypotenuse (H) for each trinagle

sin = O/H
cos = A/H
tan O/A
we have cos a = 3/5
sin a = -4 / 5
cos b = 2/3
sin b = √5 / 3
cos(a+b) =
cos a*cos b - sin a*sinb
= 3/5 * 2/3 - (-4/5)*(√5/3)
= 6/15 + 4√5 / 15
= 6 + 4√5 / 15