Help with trig type problem sin^2 theta = cos^2 theta - 1

[sydney_bristow87]

:shock: I can't do this problem... and this is only review... anyway, we were assigned two problems, and I am totally clueless as to where to even START.... I barely know what I'm solving for or why I'm solving this...

here are the problems:

sin^2 theta = cos^2 theta - 1 (solve for cos theta, limits are 0 and two pi)

and

sin^2x + cos^2x - cos x = 0 (solve for cos x and 0, limits are 0 and two pi)

 

[sydney_bristow87]

it's due tomorrow... ANY type of help at all would really be appreciated... even if you just tell me it's a trig identity or something (seriously, is it?)
I'd really like to understand the problem for once instead of the teacher looking right at me and saying "Does everyone understand it?" and then asks me if I have any questions...

 

[mathmike]

Re: Help with trig type problem sin^2 theta = cos^2 theta -

I can't do this problem... and this is only review... anyway, we were assigned two problems, and I am totally clueless as to where to even START.... I barely know what I'm solving for or why I'm solving this...

Have you ever seen this picture:

It is called trigonometry circle. It is simple and powerful guide in trigonometry world. It consists of circle drawn at the origin with radius equal to 1. You take some angle a and draw a ray from origin until it intersects the circle. Then you make two projections of that point to X axis and Y axis. The length of p rojection on Y axis is equal to sin a and projection to the X axis equals to cos a. It is directly follows from this picture and Pithagorean theorem that sin^2 a + cos^2 a = 1 (remember, that radius of the circle is 1). Angles on the cirlcle start from X axis and go counterclockwise. You can also easily see that sin(180+a) = -sin a and so forth.

Back to your original questions - try to use that main trigonometry (sin^2+cos^2=1) identity for your problems.

 

[sydney_bristow87]

Yeah, I did see some stuff like that when I googled it, but I still have no idea how to solve o.0

 

[Guest]

sin^2 theta = cos^2 theta - 1

are you sure you have this problem correctly copied as theory states this..
sin^2 theta + cos^2 theta =1
sin^2 theta = 1-cos^2 theta
which does not line up with your statement

sin^2x + cos^2x - cos x = 0 solve for cos x
look at the above statement I have written

1-cos x =0
1=cos x

 

[stapel]

Use the various trig identities you've memorized, and see about converting the equation into something that can be factored.

. . . . .sin²(@) = cos²(@) - 1

. . . . .1 - cos²(@) = cos²(@) - 1

. . . . .2 = 2cos²(@)

. . . . .1 = cos²(@)

. . . . .0 = cos²(@) - 1

. . . . .0 = [cos(@) - 1][cos(@) + 1]

Now solve the linear factors.

Apply similar techniques to the second equation.

Eliz.

 

[ChaoticLlama]

I'll start you off on the first question.

sin²Θ = 1 - cos²Θ

Unless you memorize problems, it is impossible (and not recommended) to do these problems by inspection. To solve them, use a trig identity so you have only one trig function in the equation.

First remember the trig identity:

sin²Θ + cos²Θ = 1

rearrange:

sin²Θ = 1 - cos²Θ <--- Substitute that into your original equation

There, after substitution you have:

1 - cos²Θ = cos²Θ - 1

2cos²Θ - 2 = 0

2(cos²Θ - 1) = 0

Now you start. this function equals 0 when

(cos²Θ - 1) = 0

Therefore, solve this equation, for all values when cos²Θ is equal to 1

cos²Θ = 1

have fun.

 

[soroban]

Re: Help with trig type problem sin^2 theta = cos^2 theta -

Hello, sydney_bristow87!

Are you sure those are the correct instructions?

sin²θ = cos²θ - 1. . Solve for cosθ. ? . Limits are 0 and 2pi.
.

Don't they mean: Solve for θ ?

From the identity: .sin²θ + cos²θ = 1, we have: . sin²θ = 1 - cos²θ

The equation becomes: .1 - cos²θ .= .cos²θ - 1

And we have: .2cos²θ = 2 . → . cosθ = ±1 . ---> . θ = 0, π

sin²x + cos²x - cos x = 0 . Solve for cos x and 0. ?? . Limits are 0 and 2pi.
.

Since sin²x + cos²x = 1, the equation becomes: .1 - cos x .= .0

Then we have: .cos x = 1 . ---> . x = 0

 

[sydney_bristow87]

Okay, I kind of get it... sooo I would do the second one like:
(we never took trig identities, and this is supposedly a review)

sin^2x + cos^2x - cos x = 0
+ cosx
sin^2x + cos^2x = cosx

and then i found something on google that was like sin^2 x + cos^2 x = 1

so cosx=1 so then

sin^2 x + cos^2 x -1 = 0

sin^2 x + cos^2 x = 1

x ( sin^2 + cos^2) = 0

so the answers are 1 and 0?

 

[sydney_bristow87]

yeah those are the right equations... thats what our teacher wrote on the board anyway... i honestly had no idea what he was doing, he was just like solve these for homework =/

 

[mathmike]

...
and then i found something on google that was like sin^2 x + cos^2 x = 1

so cosx=1 so then

you should've stopped right there - you almost solved the problem. If you look at the circle it is easy to see that cos x = 1 for angle x = 0, 2pi, 4pi and so on.

sin^2 x + cos^2 x -1 = 0

you can't use the equation itself to substitute into itself - you'd end up with
1=1 which is true but quite useless

x ( sin^2 + cos^2) = 0

here you made another mistake - x is the angle for which sin and cos are taken, not a number. In other words: sin^2x actually means (sin(x))^2
As such, you can't factor it out.