Is the identity I came up with for sin(x) ^ 2 correct?

[pro.in.vbdnf]

I noticed that the graphs of sin(x) and sin(x) ^ 2 are very similar. So I offset sin(x) ^ 2 to exactly match sin(x):
$$sin(x) = 2 sin^{2}\left(\frac{x}{2} +\frac{\pi}{4}\right) - 1$$

Is this right, or is it an illusion? I haven't been able to find any identity that this is based on.
If it is right, then:
$$sin^{2}(x) = \frac{sin(2x - \frac{\pi}{2}) + 1}{2}$$

Thanks,
pro.in.vbdnf

 

[arildno]

That is certainly true!

Remember that $$sin(2x-\frac{\pi}{2})=-\cos(2x)$$

and therefore, your identity is a rewriting of the half-angle formula.

 

[Elucidus]

You have indeed hit on something true. You are exploiting several trig identities. Namely:

Cofunction (for cosine): $\cos(x)=\sin\left( \frac{\pi}{2}-x \right)$

Symmetry (for sine): $\sin(-x) = \sin(x)$

Double Angle (for cosine): $\cos(2x)=1-2\sin^2(x)$

So:

$$\frac{\sin \left( 2x - \frac{\pi}{2} \right)}{2}= \frac{-\sin \left( \frac{\pi}{2} - 2x \right)}{2} = \frac{-\cos(2x)+1}{2} = \frac{-(1-2\sin^2x)+1}{2}= \frac{2\sin^2(x)}{2} = \sin^2(x)$$​

Incidently there is also the Half Angle (or Power Reducing) Identity

$$\sin^2(x) = \frac{1-\cos(2x)}{2}$$​

which is closely related to what we have here.

--Elucidus

 

[pro.in.vbdnf]

Thanks for your help! The identities I was missing putting together were $$sin^{2}(x) + cos^{2}(x) = 1$$ and $$cos(2x) = cos^{2}(x) - sin^{2}(x)$$.

So $$cos^{2}(x) = \frac{cos(2x) + 1}{2}$$. I'll have to remember that.

 

[pro.in.vbdnf]

Does that mean $$\sqrt{cos(x)}$$ has an identity in terms of cos(x)?

 

[Elucidus]

Thanks for your help! The identities I was missing putting together were $$sin^{2}(x) - cos^{2}(x) = 1$$ and $$cos(2x) = cos^{2}(x) - sin^{2}(x)$$.

So $$cos^{2}(x) = \frac{cos(2x) + 1}{2}$$. I'll have to remember that.

I believe you meant $\sin^2(x) + \cos^2(x) = 1$. The latter equation above is the Half Angle (or Power Reducing) Identity for cosine.

Does that mean $$\sqrt{cos(x)}$$ has an identity in terms of cos(x)?

I do not think so, but $$\sqrt{1 \pm \cos(x)}$$ does:

$$\sqrt{1+ \cos(x)}=\sqrt{\frac{2(1+\cos(2x/2))}{2}}=\sqrt{2\cos^2(x/2)}=\sqrt{2}\left| \cos(x/2) \right|$$

$$\sqrt{1-\cos(x)}=\sqrt{2}\left| \sin(x/2) \right|$$​

Both of these are handy for certain types of methods in calculus.

--Elucidus

 

[pro.in.vbdnf]

(Yes, I meant $\sin^2(x) + \cos^2(x) = 1$.)

I do not think so, but $$\sqrt{1 \pm \cos(x)}$$ does:

$$\sqrt{1+ \cos(x)}=\sqrt{\frac{2(1+\cos(2x/2))}{2}}=\sqrt{2\cos^2(x/2)}=\sqrt{2}\left| \cos(x/2) \right|$$

$$\sqrt{1-\cos(x)}=\sqrt{2}\left| \sin(x/2) \right|$$​

Both of these are handy for certain types of methods in calculus.

What are their names?

By the way, is there a difference between the [tex] and [itex] tags? And is there a way to vertically center the LaTeX images with the text?

 

[arildno]

1. The itex-tag can be used to generate Latex within the ordinary sentence structures, like the equation: $2x-1=3$, rather than $$2x-1=3$$

2. Note that the half-angle formula gets rid of a square root!
Thus, the right-hand side is easily integrated, the left-hand side would seem a hopeless target for integration without that nifty identity.

 

[pro.in.vbdnf]

$$\sqrt{cos(x)} = \sqrt{2 cos^{2}\left(\frac{x}{2}\right) - 1}$$
So there is no way to get rid of the right hand radical?

 

[arildno]

Sure, by squaring both sides.

 

[pro.in.vbdnf]

..while keeping the left radical? I wonder if there is an identity of the form $$\sqrt{cos(x)} = ... cos(x) ...$$?