# Homework Help: Deriving trig formulas

1. Aug 29, 2008

### 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

2. Aug 29, 2008

### 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

3. Aug 29, 2008

### Craig is Lege

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

4. Aug 29, 2008

### 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.

5. Aug 29, 2008

### danago

Try starting with cos(x + -x)

6. Aug 29, 2008

### Ben Niehoff

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

7. Aug 30, 2008

### 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.

8. Aug 30, 2008

### 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

$$f_1(-x)=-f_1(x)$$

$$f_2(-x) = f_2(x)$$

$$f_2(x + y) = f_2(x)f_2(y) - f_1(x)f_1(y)$$

$$f_1(x+y) = f_2(x) f_1(y) + f_1(x) f_2(y)$$

and it is clear that nobody will prove the identity

$$f_1^2(x) + f_2^2(x) = 1$$

now.

9. Aug 30, 2008

### atyy

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.

Last edited: Aug 30, 2008
10. Aug 30, 2008

### Defennder

? 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?

11. Aug 30, 2008

### HallsofIvy

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.

12. Aug 30, 2008

### Defennder

Ok, I see.

13. Aug 30, 2008

### tiny-tim

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.

14. Aug 30, 2008

### 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?

Carefully!

$$\cos(x) = \cos(0)\cos(x)\quad\implies\quad \cos(0)=1\;\textrm{or}\;\cos(x)=0$$

15. Aug 30, 2008

### tiny-tim

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.

16. Aug 30, 2008

### 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.

17. Aug 30, 2008

### atyy

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

18. Aug 30, 2008

### dirk_mec1

19. Aug 30, 2008

### symbolipoint

That first one , [tex] sin^2x+cos^2x=1[tex]
is merely the Pythagorean Theorem applied to the unit circle.

20. Aug 31, 2008