Finding values of sine without a calculator

How do I find the sin or inverse sin of 46 degrees (or any degree) without using a calculator?
Thanks.
Jackie

Finding an "inverse sin" doesn't apply to degrees by the way. We take sine of an angle and get a number; inverse sine is taken from a number and gives back an angle.

So is it possible to find values of sine without a calculator? I am sure there are various methods, but these two came to my mind.

1) We can go back to the definition of sine in a right triangle and using a protractor, DRAW a right triangle with that angle. Draw as accurately as you can.



Then from the picture again, we need to measure the t! wo sides: the opposite side and the hypotenuse and then calculate their ratio (paper and pencil). Again, measure as accurately as you can.

Now, as far as the opposite problem, let's say you know that the sine of some angle is 0.86 or some other number (between -1 and 1). Can we find the angle without the calculator?

Draw a right triangle with hypotenuse 1, opposite side 0.86 (or some multiple of those), and measure the angle in degrees.

This method, I feel, is a good one for demonstration and teaching purposes when first learning about sine.

2) Another possibility is to use the Taylor series of sine. Hopefully you don't need to take too many terms from it to get the desirable accuracy. This will include calculations that you'd need to do on paper.

Let's say we use Taylor series in origin and take the first four terms:
!
sin(x) ≈ x - x³/3! + x⁵/5! -! x⁷ /7!

To use this, you need to first change the 46 degrees or whatever to radians. Obviously all the calculations involved will take some time without a calculator... Auch! But an approximative method such as this that only involves the four basic operations is what your calculator probably uses, too.

Here you will find the Taylor series for inverse of sine.

3) Using sine addition formula and a known value.

Sine addition formula says:
sin(a + b) = sin a cos b + cos a sin b.

So... if you're interested in finding, say, sin(46°) and we do already happen to know sin(45°) and cos(45°)... but we need to work in radians to use the formula. So convert 46 and 45 degrees to radians (without a calculator? I'm going to cheat now...) and get 45° is Pi/4 or 0.785398, 46° is about 0.80285.

sin(Pi/4 + 0.01745) = sin(Pi/4) cos(0.01745) + cos(Pi/4) sin(0.01745).

It just so happens that for small values of x (near zero), sin x ≈ x. (You learn that in calculus, I think). So sin(0.01745) is about 0.01745. Cos(0.01745) should be pretty near 1 somewhere.

sin(Pi/4) or sin 45° is 1/√2, and cos(Pi/4) is the same. Plugging those in,

sin(Pi/4 + 0.01745) = 1/√2 * 1 + 1/√2 *0.01745.

I'm going to cheat again and do this with a calculator... to get 0.71945.

Did I get close? Well, sin46° is about 0.7193398. Got three decimals right; that's okay I guess.

Tags: sine, calculator

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Dear Korean,

The American English Professor Husband to Korean Woman and Father of a Very Cute Half Breed Kid with a Half English Half Korean Name (hereinafter referred to simply as "AEPHKWFVCHBKHEHKN") just read The Korean's blog post about learning languages. AEPHKWFVCHBKHEHKN agrees with the Korean, and gives Kudos to the hard-assed realism. People ask how AEPHKWFVCHBKHEHKN learned to speak good Arabic in just a year, and the only answer is pure hard work. (And AEPHKWFVCHBKHEHKN is a big fan of Pinker's book too.)

AEPHKWFVCHBKHEHKN has a question about language tutors. Can the Korean recommend any good tutors or language programs here in Seoul? Does the Korean himself tutor? AEPHKWFVCHBKHEHKN only has a few "free weeks" this summer, so Yonsei's lauded KSL program doesn't work for him. AEPHKWFVCHBKHEHKN asks forgiveness if this information is already on the Korean's blog.

With obsequious thanks to His Munificence, the Korean,
AEPHKWFVCHBKHEHKN

Dear AEPHKWFVCHBKHEHKN,

The Korean is glad you enjoyed the post. But he never had to learn Korean in Korea via tutoring, so he does not know at this time -- but he is sure some of AAK! readers will be able to help you. Readers, any ideas?

Got a question or a comment for the Korean? Email away at askakorean@gmail.com.

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Strategies to solve simple math equations

What are the strategies for solving simple equations?

I got this question in the mail just today.

I assume the person means LINEAR equations - those where you only have one variable (usually x), and that x is not raised to second or third or any other power, nor is it in the denominator or under square root sign or anything. Just x's multiplied by numbers and numbers by themselves, such as:

2x - 14 = 9x + 5

OR

1/3x - 3 = 2 - 1/2x

OR

2(5x - 4) = 3 + 5(-x + 1)

Here are the strategies for solving these:
* You get rid of paretheses using distributive property
* You may multiply both sides of the equation by the same number
* You may divide both sides of the equation by the same number
* You may add the same number to both sides of the equation
* You may subtract the same number from both sides of the equation

You might think, "W! hich one of those will I use, and in which order?"

That depends. There is no clear cut-n-dried answer.

Whatever you do, you try to transform your equation towards the ultimate goal: where you have x on one side alone. Also whatever you do, your goal is to transform the equation to one that you already know how to solve. It might take several steps.

For example, your first step with these equations, could be to...

1/3x - 3 = 2 - 1/2x	... multiply both sides by 6 to get rid of the fractions
2(5x - 4) = 3 + 5(-x + 1)	...multiply out the parentheses
2x - 14 = 9x + 5	...add 14 to both sides (or subtract 9x)
1/4(2x - 27 + 0.5x) = 2/5(8x + 3)	...multiply by 20 to get rid of the fractions

As with most things, practice makes perfect. ! Check also the websites below:

Tutorial on linear equations has a 4-step strategy for solving linear equations which summarizes it real well.

Algebra 1 Review - Solving Simple Equations - a step-by-step slideshow.

Ask Dr. Math ® - Solving simple linear equations - lots of examples to read here.


Tags: math, mathematics, teaching, algebra

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Dividing decimals

I feel students need to get grounded conceptually in this topic. So many times, all they learn about decimal division are the rules of how to go about decimal division when using long division, and it becomes an "empty" skill - a skill that lacks the conceptual foundation.

So for starters, we can do two different kinds of mental math division problems.

1. Division by a whole number - using mental math
   
   Here it is easy to think, "So much is divided between so many persons".
   
   0.9 ÷ 3 is like "You have nine tenths and you divide it between three people. How much does each one get?" The answer is quite easy; each person "gets" 0.3 or three tenths.
   
   And... remember ALWAYS that you can check division problems by multiplication. Since 3 × 0.3 = 0.9, we know the answer was right.
   
   0.4 ÷ 100 turns out to be an easy problem if you write 0.4 as 0.400:
   0.400 ÷ 100 is like ! "You have 400 thousandths and you divide it between 100 people; how much does each one get?" The answer is of course 4 thousandths, or 0.004. Check: 100 × 0.004 which is 100 × 4/1000 = 400/1000 or 0.400 = 0.4.
   
   Here are some more similar ones:
   
   0.27 ÷ 9
   
   0.505 ÷ 5
   
   0.99 ÷ 11
   ...and you can make more, just think of the multiplication tables.
   



2. Division where the quotient (answer) is a whole number
   
   This time it helps to think, "How many times does the divisor go into the dividend?" In these types of mental math problems, the answer ends up being a whole number. (Of course the teacher has to plan these problems just right.)
   
   For example, 0.4 ÷ 0.2. Ask, "How many times does 0.2 fit into 0.4?" The answer is, 2 times. So 0.4 ÷ 0.2 = 2. Again, we can check it by multiplying: 2 × 0.2 = 0.4.
   
   Other similar division problems to solve ment! ally:
   
   1 ÷ 0.5
   
   3 ÷ 0.5
   0.09 ÷ 0.03
   
   0.9 ÷ 0.1
   
   2 ÷ 0.4
   
   1 ÷ 0.01
   
   ...and so on.
   

This decimal division lesson taken from my Decimals 2 book illustrates these two kinds of mental division problems.

Towards the general case

After the student is familiar with the two special cases above, we can go forward and study decimal division problems in general. Even here, we will divide the problems into two classes, depending on whether the divisor is a whole number or not.

1. The divisor is a whole number.
   
   For example, 3.589 ÷ 4 or 0.1938 ÷ 83. These can simply be solved by long division as they are. Just put the decimal point in the same place in the quotient as where it is in the dividend.
   
   The "stumbling block" may come when the division is not even (this also leads into th! e study of repeating decimals). Generally, you can continue the division indefinitely by tagging zeros to the dividend, such as making 3.589 to be 3.589000. Then when you've continued the division as long as you wish (or as long as the book tells you to do it), cut the decimal off at a desired accuracy and round it.
   
   Typical problem in a textbook would say, "Do 2.494 ÷ 3 and give your answer with 3 decimal digits." For this, you need to do the long division until the fourth decimal digit - so as to be able to round to 3 decimal digits. Since 2.494 does not have four decimal digits, you tag a zero to it to make it have so (2.4940).
   
   Fortunately, this process is not generally difficult. It's the second case that's more of a problem.
   
   



2. The divisor is not a whole number.
   
   Here, we do something quite special be! fore dividing, and turn the problem into one where the divisor! is a whole number. Then, the actual division is done like explained above.
   
   I say this is special, because this special thing that we do is based on a very important general principle of arithmetic:
   
   If you multiply both the dividend and the divisor by some same number, the quotient won't change.
   
   Let's see it in action with some easy numbers:
   
   1000 ÷ 200 = 5
   
   100 ÷ 20 = 5
   
   10 ÷ 2 = 5
   
   Each time both the dividend and the divisor change by a factor of ten, but the quotient does not change.
   
   We can also try it using a factor of 3 (or any other number):
   
   8 ÷ 2 = 4
   24 ÷ 6 = 4
   72 ÷ 18 = 4
   
   Let's try one more time, with a factor of 2:
   
   30 ÷ 6 = 5
   
   15 ÷ 3 = 5
   
   7.5 ÷ 1.5 = 5
   
   3.75 ÷ 0.75 = 5
   
   H hopefully by now you have convinced the student(s) of this pri! nciple. Now we can apply it to those pesky decimal division problems.
   
   
   
   
   This image shows how the decimal division problem 0.644 ÷ 0.023 can be changed into another problem, with a whole number divisor, and with the same answer.
   
   In each step, we multiply both the dividend and the divisor by 10. This, of course, is the same process as moving the decimal point.
   
   Many textbooks only show the student the "trick" of moving the decimal point... but don't show him what that idea is based on.
   
   An example
   
   To solve 13.29 ÷ 5.19, we need to first change the problem so that the divisor 5.19 is a whole number. We multiply both the dividend and the divisor by 10 as many times as needful to accomplish that:
   
   13.29 ÷ 5.19!
   = 132.9 ÷ 51.9
   = 1329 ÷ 519, and now off ! you go t o do long division... I'm not saying it's the easiest long division problem in the world, since the divisor is 519. Let's try an easier one.
   
   
   2,916 ÷ 0.02
   = 29,160 ÷ 0.2
   = 291,600 ÷ 2 and now you can do the long division.
   
   Of course, in reality you can also multiply by 100 instead of taking two steps of multiplying by 10. But students can start out by multiplying by 10 as many times as needed.
   

Please also see the lesson on dividing decimals by decimals, from my Math Mammoth Decimals 2 book.

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