Remembering that $\sin^2(\theta) = 1/2 - 1/2\cos(2\theta)$?

Question

How do you remember this for integrals?

It doesn't seem obvious and I can never remember it when I come across it in integrals.

Answer 1

The double angles should be remembered as follows,

\begin{align*} 2\sin^2 (\theta) &= 1 - \cos(2\theta) \\ 2\cos^2 (\theta) & = 1 + \cos(2 \theta). \end{align*}

To know which sign for each, remember that 'sinning' is bad, hence it leads to $-$.

Reason why you should memorize like this is because there no fractions and you can see a pattern!

When you need to compute the integral of say, $\sin^2 (\theta)$, just recall formula and divide by $2$.

Answer 2

HINT:

I remember the identity by first remembering the addition formula:

$$\cos (x+y)=\cos x \cos y-\sin x \sin y \implies \cos (2x) =1-2\sin^2 x$$

Answer 3

Its easy to remember cos(2x) = cos²x - sin²x and then use cos²x + sin²x = 1 to get the required identity

Answer 4

here is a geometric way to remember the double angle formulae. we will use the fact that on the unit circle the terminal point corresponding to the arc length $t$ is $(\cos t, \sin t).$

look at the arc of length $2t$ starting at $A = (1,0)$ and ending at $B = (\cos 2t , \sin 2t)$ the midpoint $C$ of $AB$ has coordinate $((1+\cos 2t)/2, \sin 2t/2).$ but the length $OC$ is $\cos t.$

now the point on the unit circle cut by $OC$ can be seen in two ways:

$$a: (\cos t, \sin t), \quad b: \left(\frac{1+\cos 2t}{2\cos t}, \frac{\sin 2t}{2\cos t}\right)$$

equation the two gives the double angle formula $$1+\cos 2t = 2\cos^2 t, \sin 2t = 2\sin t \cos t $$

Answer 5

$\cos^2 \theta + \sin^2 \theta = 1$
$\cos^2 \theta - \sin^2 \theta = \cos 2\theta$

Add and you get
$2\cos^2 \theta = 1 + \cos 2\theta$
$\cos^2 \theta = \frac 1 2 + \frac 1 2\cos 2\theta$

Subtract and you get
$2\sin^2 \theta = 1 - \cos 2\theta$
$\sin^2 \theta = \frac 1 2 - \frac 1 2\cos 2\theta$

Answer 6

Here's a longer explanation, which maybe will make the identity seem a bit more obvious:

As you know, a cosine wave oscillates between $-1$ and $+1$ with a period of $2 \pi$. Here are some of the values: $$ \cos 0=+1 ,\quad \cos(\pi/2) = 0 ,\quad \cos\pi = -1 ,\quad \cos(3\pi/2) = 0 ,\quad \cos(2\pi) = +1 . $$ Compare this to the values you get if you square everything: $$ \cos^2 0=+1 ,\quad \cos^2(\pi/2) = 0 ,\quad \cos^2\pi = +1 ,\quad \cos^2(3\pi/2) = 0 ,\quad \cos^2 2\pi = +1 . $$ So a cosine-squared wave oscillates between $0$ and $+1$, and it makes two oscillations where the cosine wave makes only one, so it oscillates twice as fast.

In fact, it's a perfect sin/cos-wave shaped oscillation. Of course, this isn't obvious just from sampling a few values, but it's true; that's what the double-angle formula is saying. Try plotting it on a computer, and you'll see that it certainly looks that way!

So it must be oscillating around its average value $1/2$ with an amplitude of $1/2$ (so that it precisely reaches up to its maximum value $1$ and reaches down to its minimum value $0$).

Thus, you get the cosine-squared wave by taking a cosine wave $\cos 2\theta$ (with twice the frequency compared to $\cos \theta$), multiplying it by the amplitude factor $1/2$, and then adding $1/2$ to shift the graph upwards: $$ \cos^2 2 \theta = \frac12 + \frac12 \cos 2\theta . $$ And the formula for the sine-squared that you asked about is exactly similar; it's just that you multiply by the amplitude factor $-1/2$ instead, to flip the wave upside down. (Since it starts out with $\sin^2 0 = 0 = 1/2 - 1/2$ instead of $\cos^2 0 = 1 = 1/2 + 1/2$.)

Answer 7

If I want to recall the formulas $|\sin \frac x2| = \sqrt{\frac{1-\cos x}2}$ and $|\cos \frac x2| = \sqrt{\frac{1+\cos x}2}$ and if I already remember the form $$\sqrt{\frac{1\pm\cos x}2}$$ but I do not remember which sign corresponds to sine and which to cosine, I can recall it like this:

If $x$ is close to $0$, then $\cos x$ is close to $1$. I want to get the value of $\sin(x/2)$, which is close to $0$, which I get from $1-\cos x$. Similarly, I want to get $\cos(x/2)$ close to $1$, to get this, I take $1+\cos x$.

But I agree that probably the better way is to recall how the formulas were derived from the double angle formulas.