# Thread: trig (obviously)

1. ## trig (obviously)

find the solution for the equation tanA=3sinA for values of A in the range -180 <= A <= 180 (deg evidently)
so i used tanA= sinA/cosA which gives cosA = sinA/3sinA = 1/3. anything wrong up until here?
my calculator says cos^-1 1/3 = 70.5 deg, so my answer is +- 70.5 deg.
however the answer includes +-180° as well. is this even possible? who is wrong?

2. Originally Posted by furor celtica
find the solution for the equation tanA=3sinA for values of A in the range -180 <= A <= 180 (deg evidently)
so i used tanA= sinA/cosA which gives cosA = sinA/3sinA = 1/3. anything wrong up until here?
my calculator says cos^-1 1/3 = 70.5 deg, so my answer is +- 70.5 deg.
however the answer includes +-180° as well. is this even possible? who is wrong?
The book is right. Note that $\tan A=3\sin A\implies 3\sin A-\tan A=0\implies 3\sin A-\frac{\sin A}{\cos A}=0$ $\implies \sin A\left(3-\frac{1}{\cos A}\right)=0$

So $\sin A=0$ or $\cos A=\frac{1}{3}$. That's why $\pm180^{\circ}$ is included in the solution set. (It seems like one more value is missing from the book's answer...can you see what it is?)

Does this clarify things?

3. but how could i know this just by using my method?

4. Originally Posted by furor celtica
but how could i know this just by using my method?

Chris has shown it to you . However ,

$\ tanA=3 \ sinA$

$\frac{\ sinA}{\ cosA}=3\ sinA$

$\ sinA=3\ sinAcosA$

$3\ sinAcosA-\ sinA=0$

$\ sinA(3\ cosA-1)=0$

$\ sinA=0$ or $\ cosA=\frac{1}{3}$