# Tag Info

1

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 $1$. It want to ...

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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 ...

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

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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 ...

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

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HINT: I remember the identity by first remembering the addition formula: $$\cos (x+y)=\cos x \cos y-\sin x \sin y \implies \cos (2x) =1-2\sin^2 x$$

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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 ...

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