Deriving Trig Formulas: Solving Problem with Sin and Cos Identities

[Craig is Lege]

hello. i need help deriving trig formulas.

the first problem i most derive is

51. sin^2 x + cos^2 x = 1


i can use these identities:

a. sin(-x) = -sinx
b. cos(-x) = cos(x)
c. cos(x+y) = cosxcosy - sinxsiny
d. sin(x+y) = sinxcosy + cosxsiny

thanks for any help

 

[Craig is Lege]

hello. i need help deriving trig formulas.

the first problem i most derive is

51. sin^2 x + cos^2 x = 1


i can use these identities:

a. sin(-x) = -sinx
b. cos(-x) = cos(x)
c. cos(x+y) = cosxcosy - sinxsiny
d. sin(x+y) = sinxcosy + cosxsiny

thanks for any help

 

[Craig is Lege]

i think i posted in the wrong area...sorry...my bad

 

[sutupidmath]

well, the identity

sin^2x+cos^2x=1 is the fundamental identity in trig. It can be derived directly from the unit trig circle.

So, what u basically do is this. You draw the oriented trig. circle, which means that your radius is 1. then your sinx will lay on the vertical line ( along the y-axis, if you place the trig circle in the origin of the coordinate system) and your cosx will lie along the x-axis.

Now you need to use pythagorean theorem, so you'll have :x^2+y^2=1 but remember that x is actually the sin and y cos. and radius is 1.

 

[danago]

Try starting with cos(x + -x)

 

[Ben Niehoff]

Look at your equation C. What happens when you set y = -x?

 

[Gib Z]

It's ok, at least you know now.

I don't actually believe its possible to derive that equation 51 from those listed identities. It's much easier to derive using the definition of the trig functions on the unit circle anyway - much more fundamental. Identities a and b are quite basic and don't exactly help, and c & d are more complex identities than eqn 51 itself.

 

[jostpuur]

There is indeed some unclarity about what one is supposed to use in the proof. If we define two functions, both identically zero, f_1=0 and f_2=0, then these functions satisfy the properties

<br /> f_1(-x)=-f_1(x)<br />

<br /> f_2(-x) = f_2(x)<br />

<br /> f_2(x + y) = f_2(x)f_2(y) - f_1(x)f_1(y)<br />

<br /> f_1(x+y) = f_2(x) f_1(y) + f_1(x) f_2(y)<br />

and it is clear that nobody will prove the identity

<br /> f_1^2(x) + f_2^2(x) = 1<br />

now.

 

[atyy]

How about using your third identity, with y=-x:

LHS: cos(x+(-x)) = cos(0) = 1

RHS: cos(x)cos(-x) - sin(x)sin(-x)
=cos(x)cos(x) -sin(x)[-sin(x)] (Here we use your first two identities)
=cos²x+sin ²x

So the identities you gave are not enough. We also needed cos(0)=1.

 

[Defennder]

There is indeed some unclarity about what one is supposed to use in the proof. If we define two functions, both identically zero, f_1=0 and f_2=0, then these functions satisfy the properties

<br /> f_1(-x)=-f_1(x)<br />

<br /> f_2(-x) = f_2(x)<br />

<br /> f_2(x + y) = f_2(x)f_2(y) - f_1(x)f_1(y)<br />

<br /> f_1(x+y) = f_2(x) f_1(y) + f_1(x) f_2(y)<br />

and it is clear that nobody will prove the identity

<br /> f_1^2(x) + f_2^2(x) = 1<br />

now.

? I don't see what this has got to do with the original post. Your f1 and f2 are constant zero functions, so how does this relate to sin and cos?

 

[HallsofIvy]

? I don't see what this has got to do with the original post. Your f1 and f2 are constant zero functions, so how does this relate to sin and cos?

His point is that f1 and f2 satisfy all the given equations except f1²+ f2²= 1 so it is impossible to use the given equations to prove that.

 

[Defennder]

Ok, I see.

 

[tiny-tim]

How about using your third identity, with y=-x:

LHS: cos(x+(-x)) = cos(0) = 1

RHS: cos(x)cos(-x) - sin(x)sin(-x)
=cos(x)cos(x) -sin(x)[-sin(x)] (Here we use your first two identities)
=cos²x+sin ²x

Hi atyy! Nice!

So the identities you gave are not enough. We also needed cos(0)=1.

No … we can say sin(0) = sin(-0) = -sin(0), so sin(0) = 0.

Then cos(x) = cos(0 + x) = cos(0)cos(x) + 0 sin(x), so cos(0) = 1.

 

[jostpuur]

atyy, very nice remark. I didn't notice that such small addition to the assumptions was enough. btw. the original post was quite hmhm.. "homework like". Could it be that that was a little bit too explicit answer?

Hi atyy! Nice!


No … we can say sin(0) = sin(-0) = -sin(0), so sin(0) = 0.

Then cos(x) = cos(0 + x) = cos(0)cos(x) + 0 sin(x), so cos(0) = 1.

Carefully!

<br /> \cos(x) = \cos(0)\cos(x)\quad\implies\quad \cos(0)=1\;\textrm{or}\;\cos(x)=0<br />

 

[tiny-tim]

Carefully!

<br /> \cos(x) = \cos(0)\cos(x)\quad\implies\quad \cos(0)=1\;\textrm{or}\;\cos(x)=0<br />

No … \implies\quad \cos(0)=1\;\textrm{ , or}\;\cos(x)=0 \text{ FOR ALL x} …

and if cosx = 0 for all x, then sinx = 0 also.

 

[jostpuur]

I see. So we don't need to assume that cos(0)=1, but it suffices to assume that cos(x)!=0 with some x, and then cos(0)=1 follows.

 

[atyy]

There is indeed some unclarity about what one is supposed to use in the proof. If we define two functions, both identically zero

we can say sin(0) = sin(-0) = -sin(0), so sin(0) = 0.

Then cos(x) = cos(0 + x) = cos(0)cos(x) + 0 sin(x), so cos(0) = 1.

Carefully!

<br /> \cos(x) = \cos(0)\cos(x)\quad\implies\quad \cos(0)=1\;\textrm{or}\;\cos(x)=0<br />

No … \implies\quad \cos(0)=1\;\textrm{ , or}\;\cos(x)=0 \text{ FOR ALL x} …

and if cosx = 0 for all x, then sinx = 0 also.

Wow, jostpuur and tiny-tim! That's the nicest thing I've learned in some time.

 

[dirk_mec1]

see here:

https://www.physicsforums.com/showthread.php?t=252601

 

[symbolipoint]

That first one , sin^2x+cos^2x=1&lt;br /&gt; is merely the Pythagorean Theorem applied to the unit circle.

 

[Craig is Lege]

i made progress, but hit a dead end again...

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

can anyone bring it home...

 

[Defennder]

see here:

https://www.physicsforums.com/showthread.php?t=252601

You provided a link to this thread itself.

i made progress, but hit a dead end again...

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

can anyone bring it home...

Didn't you follow what was written by the others earlier? And that isn't correct, since cos(-x)cos(x) is not 0. Furthermore, why did you start with sin^2 x + cos^2 x = 1? That is what you're trying to prove, right?

 

[HallsofIvy]

i made progress, but hit a dead end again...

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

can anyone bring it home...

You have been told repeatedly that you CAN'T prove sin²(x)+ cos²(x)= 1 from those other properties and Jostpuur showed why: two identically zero functions will satify all the other properties but not the Pythagorean identity.