Thru what method would you solve this problem?

cos^2x = sin^2x +1

Linear Method

Factoring

Quadratic Formula

Squaring :D

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- May 13th 2005, 08:32 PMjp25111Trigonometric Equations
Thru what method would you solve this problem?

cos^2x = sin^2x +1

Linear Method

Factoring

Quadratic Formula

Squaring :D - May 14th 2005, 04:58 AMticbol
Are we limited to those four methods?

If not, then I will use substitution as the first step, and then I will see the resulting equation. It could then be by factoring or by using the quadratic formula, or by whatever.

cos^2x = sin^2x +1

Rewriting that,

cos^2(x) = sin^2(x) +1 ----(i)

First, I will use the trig identity

sin^2(x) +cos^2(x) = 1

to transform Eq.(i) into sines only.

From that trig identity,

cos^2(x) = 1 -sin^2(x)

Substitute that into (i),

1 -sin^2(x) = sin^2(x) +1

Collect like terms,

-sin^2(x) -sin^2(x) = 1 -1

-2sin^2(x) = 0

Divide both sides by -2,

sin^2(x) = 0

Take the square roots of both sides,

sin(x) = 0

Take the arcsine of both sides,

x = arcsin(0) ----***

If we limit the domain of x from 0 to 360 degrees only,

x = 0, 180, or 360 degrees. ----answer.

---------------

Another way.

We transform Eq.(i) into cosines only.

From the same trig identity we used above,

sin^2(x) = 1 -cos^2(x)

Substitute that into (i),

cos^2(x) = [1 -cos^2(x)] +1

cos^2(x) = 1 -cos^2(x) +1

cos^2(x) +cos^2(x) = 1 +1

2cos^2(x) = 2

Umm, can be by factoring.

2cos^2(x) = 2

2cos^2(x) -2 = 0

2(cos^2(x) -1) = 0

Divide both sides by 2,

cos^2(x) -1 = 0

That quadratic equation is a difference of two squares,

factoring that,

(cos(x) +1)*(cos(x) -1) = 0

cos(x) +1 = 0

cos(x) = -1

x = arccos(-1)

x = 180 degrees

cos(x) -1 = 0

cos(x) = 1

x = arccos(1)

x = 0, or 360 degrees

Therefore, if x is from 0 to 360 degrees only, then,

x = 180, 0, or 360 degrees. ---same as in the first way above. - May 14th 2005, 09:04 AMhpeQuote:

Originally Posted by**jp25111**

cos^2(x) - sin^2(x) = 1

cos(2x) = 1

so 2x = 0 or 2pi or 4pi ...

so x must be a multiple of pi - as in ticbol's solution.

Check whether this always gives a solution (always a good idea for trig. equations):

cos (k*pi) = +1 or -1, sin(*pi) = 0,

so these are indeed solutions of the problem. - May 17th 2005, 09:10 AMjp25111
You guys are awesome. Thanks soooooooo much!