I have a test tomorrow and I am having trouble remembering those pesky trigonometrical identities (such as $1-\cos x=2\sin^2(\frac{x}{2})$ )
Do you guys have any tips on how I can remember these?
Thanks :)
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$\begingroup$For what it's worth:
I just memorize one Pythagorean identity and one of the sum identities. Many of the others (besides the obvious ones: the reciprocal, periodicity, and Pythagorean) can be derived starting with one of the sum formulas.
So, you could just memorize how to derive them. Of course, in a test scenario, this may waste precious time...
Reciprocal identities
The reciprocal identities follow from the definitions of the trigonometric functions.
$$\eqalign { \sec\theta&= {1\over \cos\theta} \qquad \tan\theta= {\sin\theta\over \cos\theta} \cr \csc\theta&= {1\over \sin\theta} \qquad \cot\theta= {1\over \tan\theta} \cr } $$
Periodicity relations
The Periodicity relations follow easily by considering the involved angles on the unit circle.
$$\def\ts{}\eqalign { \sin(\theta)&= \sin(\theta \pm2k\pi) \qquad \csc(\theta)= \csc(\theta \pm2k\pi) \cr \cos(\theta)&= \cos(\theta \pm2k\pi) \qquad \sec(\theta)= \sec(\theta \pm2k\pi)\cr \tan(\theta)&= \tan(\theta \pm k\pi)\phantom{2} \qquad \cot(\theta)= \cot(\theta \pm k \pi) \cr } $$
$$\eqalign { \sin(\theta)&= - \sin(\theta -\pi) \qquad \csc(\theta)= - \csc(\theta -\pi) \cr \cos(\theta)&= - \cos(\theta -\pi) \qquad \sec(\theta)= - \sec(\theta -\pi) \cr \tan(\theta)&= - \tan(\theta -\ts{\pi\over2}) \qquad \kern-3pt \cot(\theta)= - \cot(\theta -\ts{\pi\over2}) \cr } $$
Pythagorean Identities
The first Pythagorean Identity follows from the Pythagorean Theorem (look at the unit circle). The other two Pythagorean Identities follow from the first by dividing both sides by the appropriate expression (divide through by $\sin$ or by $\cos$ to obtain the other two).
$$\eqalign { \sin^2\theta +\cos^2\theta&=1\cr 1+ \cot^2\theta& =\csc^2\theta\cr \tan^2\theta + 1& = \sec^2\theta} $$
Sum and difference formulas
Memorize the first sum and difference formula. The second one can be derived from the first using the fact that $\sin$ is an odd function.
One can then derive the last two sum identities by using the first two and the fact that $\cos(\theta-\pi/2)=\sin\theta$.
$$\eqalign{ \cos(x+y)&=\cos x\cos y-\sin x\sin y\cr \cos(x-y)&=\cos x\cos y+\sin x\sin y\cr \sin(x+y)&=\sin x\cos y+\sin y\cos x\cr \sin(x-y)&=\sin x\cos y-\sin y\cos x\cr } $$
Double angle formulas
The Double Angle formulas for $\sin$ and $\cos$ are derived by using the Sum and Difference formulas by writing, for example $\cos(2\theta)=\cos(\theta+\theta)$ and using the Pythagorean Identities for the $\cos$ formula (I suppose the formula for $\tan$ should be memorized).
$$\eqalign{ \sin(2\theta)&=2\sin\theta\cos\theta \cr \tan(2\theta)&= {2\tan \theta\over 1-\tan^2\theta } \cr \cos(2\theta)&= \cos^2\theta-\sin^2\theta \cr &=2\cos^2\theta -1\cr &=1-2\sin^2\theta\cr } $$
Half angle formulas
The Half-Angle formulas for $\sin$ and $\cos$ are then obtained from the Double Angle formula for $\cos$ by writing, for example, $\cos\theta=\cos(2\cdot{\theta\over2})$
The $\tan$ formula here can easily be obtained from the other two. (Note the forms for the $\cos$ and $\sin$ formulas. These aren't to hard to memorize)
$$\eqalign{ \cos{\theta\over2}&= \pm\sqrt{1+\cos\theta\over2}\cr \sin{\theta\over2}&= \pm\sqrt{1-\cos\theta\over2}\cr \tan{\theta\over2}&=\pm\sqrt{1-\cos\theta\over1+\cos\theta} }$$
$\endgroup$ 4 $\begingroup$My favourite trick: I don't remember any of them. :-) The only thing I have in mind is that this matrix
$$ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} $$
rotates vectors in the plane by an angle $\theta$ and matrix multiplication is the same as composition. Hence, you have identities like
$$ \begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} $$
from which it follows
$$ \cos(2\theta) = \cos^2\theta - \sin^2\theta $$
and
$$ \sin(2\theta) = 2\sin\theta\cos\theta \ . $$
Alternatively, as yoyo says, you could use Euler's identity,
$$ e^{i\theta} = \cos\theta + i \sin\theta $$
to find, for instance, that
$$ \cos(\theta + \phi) + i\sin(\theta + \phi) = e^{i(\theta + \phi)} = e^{i\theta}e^{i\phi} = (\cos\theta + i\sin\theta) (\cos\phi + i\sin\phi) \ . $$
Hence,
$$ \cos(\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi $$
and
$$ \sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi \ . $$
$\endgroup$ 4 $\begingroup$you should remember the following $$ \sin(a\pm b)=\sin(a)\cos(b)\pm\cos(a)\sin(b) $$ $$ \cos(a\pm b)=\cos a\cos b\mp\sin a\sin b $$ $$ \cos^2x+\sin^2x=1 $$ from these (with $x=a=b$) you can get $$ \cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x=2\cos^2x-1 $$ and various other rearrangements.
this may not be helpful, but when I couldn't remember sine/cosine of a sum (ie before teaching) i would use $$e^{i\theta}=\cos\theta+i\sin\theta$$ and $$ e^{i(\theta+\phi)}=e^{i\theta}e^{i\phi} $$ to find them (or check my memory)
$\endgroup$ 1 $\begingroup$Apart from the three sum/difference formulas posted by yoyo which are useful enough to know by heart use the visualization of a moving vector with endpoint on the unit circle. The x-coordinate is cos(a), the y-coordinate is sin(a), a beeing the angle measured counterclockwise from the x-axis.
Most formulas can be derived from this small set of tools fairly easily.
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