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1

My 4th grade teacher had a story about a guy who was so dumb he was given a test to fill out: 2x9= 3x9= 4x9= 5x9= 6x9= 7x9= 8x9= 9x9= The guy was utterly dumb and had no idea what to do. So he just wrote the numbers 1,2,3 down the paper until he reached the bottom. The he started from the bottom and wrote the numbers 1,2,3 up the paper until he ...


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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 = ...


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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 ...


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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 ...


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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 ...



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