trig (obviously)

[red and white kop!]

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 :evil: ? who is wrong?

 

[arthur ohlsten]

tanA=3 sinA -180<A<180

solution 1
tanA=3 sinA=0
A= 180 or -180 answer


solution 2
sinA/CosA =3 sin A
...
A=+/- 70.5

Arthur

 

[wjm11]

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.

Hi, red and white,

Here is how to see what is going on: graph two functions, y = tanA and y = 3sinA on the same set of axes. Graph from x = -180 to x = 180. The points where the two functions intersect are your solutions.

If you follow this general approach, it will help you find solutions to similar problems in the future.

 

[Mrspi]

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?<-------YES, something is wrong. When you divided both sides by sin A, you're making the assumption that sin A is NOT EQUAL TO 0....thus you eliminated +/- 180 as a solution
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 :evil: ? who is wrong?

You've substituted (sin A / cos A) for tan A....

sin A / cos A = 3 sin A

It's clear that cos A cannot equal 0, because if it were 0, the fraction would be undefined. So, it's safe to multiply both sides of the equation by cos A:

sin A = 3 sin A cos A

Now, subtract 3 sin A cos A from both sides:

sin A - 3 sin A cos A = 0

Factor sin A out of both terms on the left side:

sin A (1 - 3 cos A) = 0

Set each factor equal to 0 and solve...

if sin A = 0, then A = +180 or -180

if 1 - 3 cos A = 0, then cos A = 1/3 and that gives you the other solutions you stated.

 

[Deleted member 4993]

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 :evil: ? who is wrong?

I think both (you and the book)are wrong!

Replace A by 180° and you'll get

tan(180°) = 0

3sin(180°) = 0

So at A = 180°, the equation holds good and thus A = 180° is a solution.

Similarly, at A = -180°, the equation holds good and thus A = ±180° are solutions.( Book is correct here)

The best way to these types of problems is to leave "canceling" for later step.

tan(A) = 3 sin(A)

We know cos(A) is not equal to zero (because A = 90° is not a solution)

so we can multiply both sides by cos(A).

tan(A) * cos(A) = 3 sin(A) * cos(A)

sin(A) = 3 sin(A) * cos(A)

sin(A) - 3 sin(A) * cos(A) = 0

sin(A) * [1 - 3 cos(A)] = 0

then either

sin(A) = 0 ---> A = ±180° and 0°

or

cos(A) = 1/3 ---> A = 70.5°

Now -- I wonder why the book did not include 0° as a solution??!! Or did you miss it?

 

[red and white kop!]

good point. no, the book did not include 0°, so i'm not the only one wrong here, na na na :lol: