# Thread: Trigonometric Equations - Different Methods/Different Results

1. ## Trigonometric Equations - Different Methods/Different Results

This is the equation:

1 + Sin(theta) = 2 Cos squared(theta)

The method I used goes like this:

1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities

1 + Sin(theta)
------------------------------ = 2 divide by [1 - Sin squared(theta)]
(1 + Sin(theta)) (1 - Sin(theta))

1
--------------- = 2
(1 - Sin(theta))

1 - Sin(theta) = 1/2

-Sin(theta) = -1/2

Sin(theta) = 1/2

theta = PI/6 + 2k(PI) or theta = 5PI/6 + 2k(PI)

restrict the solution to 0 < theta < 2PI

therefore the solution is {PI/6, 5PI/6}

HOWEVER you get a MORE complete solution if you work via this method:

1 + Sin(theta) = 2 Cos squared(theta) original equation

1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities

1 + Sin(theta) = 2 - 2 Sin squared(theta) multiplies out the right side

2 Sin squared(theta) + Sin(theta) - 1 = 0 rearranges the equation

(2 Sin(theta) - 1) (Sin(theta) + 1) = 0 factors left side

2 Sin(theta) - 1 = 0 or Sin(theta) + 1 = 0

Sin(theta) = 1/2 or Sin(theta) = -1

theta = {PI/6, 5PI/6} or theta = 3PI/2

Clearly the second method yields a more complete answer, and I verified it with a TI-89. Yet the first method is algebraicly sound, as far as I can tell. Any thoughts on this?

2. Because you cannot divide by,
$1-\sin^2 \theta$!!!!

Because it might be zero!
Then you divided by zero!
(Oh no!)

Once you divide by zero you get strange results.
That was you fault.
-----

$1+\sin x=2(1-\sin^2 x)$
Thus,
$(1+\sin x)-2(1+\sin x)(1-\sin x)=0$
Factor,
$(1+\sin x)[1-2(1-\sin x)]=0$
Thus,
$(1+\sin x)(2\sin x-1)=0$
Now set each factor equal to zero.

3. Originally Posted by spiritualfields
This is the equation:

1 + Sin(theta) = 2 Cos squared(theta)

The method I used goes like this:

1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities

1 + Sin(theta)
------------------------------ = 2 divide by [1 - Sin squared(theta)]
(1 + Sin(theta)) (1 - Sin(theta))
This is only valid when both $1 + \sin(\theta) \ne 0$ and $1 - \sin(\theta) \ne 0$, so these cases need to be checked separately.

RonL

4. Thank you, The Perfect Hacker and Captain Black. Interestingly, with this particular equation, the value that did not show up when I divided by (1 + Sin(theta)) (1 - Sin(theta)) was 3PI/2, which in fact is the value that creates the divide by zero situation for this equation ... (1 + Sin(3PI/2)).

Ed