Tangent Identities

[giraffie]

tan a=3/4, tan b=-3/4, a is in the first quadrant, and b is in the second. Find sin (a+b)

To be honest, I don't even know where to start. I figured that angle b had to have coordinates (-root three/two, root three/2), and the same but with no negative for a. Is that correct? And what should my next step or two be?

 

[Loren]

Draw a sketch of a and another of b. You say a is in QI. So, the sketch will be a right triangle with the hypotenuse going from (0,0) to (4,3). One leg is from (0,0) to (4,0) and the other leg is from (4,0) to (4,3). Now, use the Pythagorean Thm to calculate the measure of the hypotenuse. In like manner, label your sketch of angle b in QII. After you have all the sides labeled use the identity sin(a+b) = sin a cos b + cos a sin b to calculate the value of sin(a+b).

In answer to your statement "I figured that angle b had to have coordinates (-root three/two, root three/2), and the same but with no negative for a. Is that correct?" --- angles don't have coordinates. The vertices of angle b are at (0,0), (-4,0) and (-4,3).

 

[soroban]

Hello, giraffie!

\(\displaystyle \tan a = \tfrac{3}{4},\;\tan b = -\tfrac{3}{4}\)
\(\displaystyle a\) is in quadrant 1, \(\displaystyle b\) is in quadrant 2.

\(\displaystyle \text{Find }\:\sin(a+b)\)

\(\displaystyle \text{We have: }\:\tan a \:=\:\tfrac{3}{4} \:=\:\tfrac{opp}{adj}\)

\(\displaystyle a\text{ is in quadrant 1 with: }opp = 3,\;adj = 4\)
. . \(\displaystyle \text{Pythagorus says: }\:hyp = 5\)
\(\displaystyle \text{Hence: }\:\sin a = \tfrac{3}{5},\quad \cos a = \tfrac{4}{5}\) .[1]


\(\displaystyle \text{We have: }\:\tan b \:=\:\text{-}\tfrac{3}{4} \:=\:\tfrac{opp}{adj}\)

\(\displaystyle b\text{ is in quadrant 2 with: }opp = 3,\;adj = \text{-}4\)
. . \(\displaystyle \text{Pythagorus says: }\:hyp = 5\)
\(\displaystyle \text{Hence: }\;\sin b = \tfrac{3}{5},\quad\cos b = \text{-}\tfrac{4}{5}\) .[2]


\(\displaystyle \text{We know that: }\;\sin(a + b) \;=\;\sin a \cos b +\cos a\sin b\) .[3]


Substitute [1] and [2] into [3]:

. . \(\displaystyle \sin(a + b) \;=\;\left(\frac{3}{5}\right)\left(\text{-}\frac{4}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{3}{5}\right) \;=\;\text{-}\frac{12}{25} + \frac{12}{25} \;=\;0\)