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Multiplying by $9$ is relatively easy, since $9a=10a-a$. I.e., multiply by ten and then subtract. (And multiplication by ten is just adding zero at the end.) Examples: $6\times 9 = 6\times 10 - 6 = 60 - 6 = 54$ $12\times 9 = 12\times 10 - 12 = 120 - 12 = 108$ $16\times 9 = 16\times 10 - 16 = 160 - 16 = 144$ $24\times 9 = 24\times 10 - 24 = 240 - 24 = ... 0 GEMS, for order of operations. Groupings, Exponents, Multiplication/division, subtraction, far superior to PEMDAS. A little more advanced is the function composition mnemonic: If I put on my socks and then my shoes, when I compute the inverse, I take off my shoes and then my socks. I always thought it was funny because it took me longer to understand the ... 0 Right away, you can cross off the fourth formula, since it is equivalent to the third formula after switching$a$and$b$. Then, you can also avoid the last four formulas, since these are all covered by the first three formulas via the relationships $$a+b = u, \quad a-b = v, \quad a = \frac{u+v}{2}, \quad b = \frac{u-v}{2}.$$ So that really leaves us with ... 0 How about just restating the LHS. For example, you could restate$\cos a\sin b$as $$\frac{\sin a\cos b +\cos a\sin b + \cos a\sin b - \sin a\cos b}{2}$$ and just figure it out from there. 5 The only ones you need to know are the classical$\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$and$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. The others are mere consequences of those. For example, by changing the signs, you get$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$. By summing, you have$\cos(a+b)+\cos(a-b) = 2\cos(a)\cos(b)\$, which is your first ...