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5

I find that four suffice. $$\cos^2 (x) + \sin^2(x) = 1 \tag{1}$$ $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) \tag{2}$$ $$\sin(a+b)=\sin(a)cos(b)+\sin(b)\cos(a) \tag{3}$$ $$trig(x) = cotrig(\frac{\pi}{2}-x) \tag{4}$$ The Pythagorean identity $(1)$ is easy to manipulate. Divide through by $cos^2(x)$ alternatively by $sin^2(x)$ to find the other ...


4

From the comments to the question: Michael Burr: The definitions of $\tan$, $\csc$, $\sec$, and $\cot$ must be memorized (mnemonic: each pair of reciprocals has one "co" e.g., $(\sin,\csc)$ are a pair) I memorize $\sin^2(x)+\cos^2(x)=1$ and the angle sum/difference formulas. Everything else can usually be derived from those. This is great ...


1

A systematic way to implement what pbs wrote, is mentioned in the book A = B in section 1.5: Let $w = \exp(ix)$, then $\cos x = (w +w^{-1})/2$ and $\sin x = (w - w^{-1}/(2i$). So equality of rational expressions in trigonometric functions can be reduced to equality of polynomial expressions in w. (Exercise: Prove, in this way, that $\sin 2x = 2\sin ...


14

I always found recalling $e^{ix}=\cos x+i\sin x$ useful for quickly deriving the sum of angles formulae, e.g. $$e^{i(x+y)}=\cos(x+y)+i\sin(x+y).\tag{1}$$ But $$e^{ix+iy}=e^{ix}e^{iy}=(\cos x+i\sin x)(\cos y+i\sin y).\tag{2}$$ Expanding (2), equating with (1) and separating real and imaginary parts gives you the formulae. You can then get the double ...


1

Stan's answer will probably do a better job than mine; while running the risk of repeating information: memorize $\sin^2(x)$ $+$ $\cos^2(x)$ $=$ $1$, dividing through by $\cos^2(x)$ will yield the identity $\tan^2(x)$ + $1$ $=$ $\sec^2(x)$. Additionally, dividing through by $\sin^2(x)$ will yield $1 +$ $\cot^2(x)$ = $\csc^2(x)$. Furthermore, intuitively ...



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