Typically, when dealing with trig functions, we see “nice” values for sines and cosines… and by that I mean it’s possible to obtain exact values for them. For example, stuff like π/6, π/4, π/3, π/2, or multiples thereof. But what about the the expression is a value we can’t obtain an exact value for? For example, what consider the following:

cos(2π/7) +cos(4π/7) + cos(8π/7) = -1/2

How in the world can this be proved (or not)? We don’t even have a value for cos(π/7), let alone most multiples of it. What do we do?

The best way to prove anything will be to see of we can get terms to cancel. That is, can we reduce the complexity of the equation down to something that’s always true? For example, cos(π/7)= cos(π/7)? Or even something that’s not true, such as cos(3π/7)=0. However we can get to something like this, that will prove – or disprove – our equation.

That said, let’s try to simplify the left side. In other words, we want to be able to try and collect like terms.

For starters, there is a trigonometric identity that says:

sin(x)cos(y) = (1/2)[(sin(x+y)+sin(x-y)]

Let's multiply both sides of the equation by sin(π/7):

sin(π/7)cos(2π/7) + sin(π/7)cos(4π/7) + sin(π/7)cos(6π/7) = -1/2

Now, we apply the identity above:

(1/2)[sin(π/7+2π/7) + sin(π/7-2π/7)] + (1/2)[sin(π/7+4π/7) + sin(π/7-4π/7)]

+ (1/2)[sin(π/7+8π/7) + sin(π/7-8π/7)] = -(1/2)sin(π/7)

We can multiply both sides by 2 and simplify the terms a bit:

[sin(3π/7) + sin(-π/7)] + [sin(5π/7) + sin(-3π/7)] + [sin(9π/7) + sin(-7π/7)] = -sin(π/7)

Further simplifying:

sin(3π/7) - sin(π/7) + sin(5π/7) - sin(3π/7) + sin(9π/7) - sin(7π/7) = -sin(π/7)

Now, we can collect like terms:

sin(5π/7) + sin(9π/7) - sin(7π/7) = 0

But sin(7π/7) = 0:

sin(5π/7) + sin(9π/7) = 0

That’s a lot simpler. But what does sin(5π/7) + sin(9π/7) equal, anyway? Do they sum to zero?

Maybe the following diagram will help:

Each point has the coordinates (sin<angle>, cos<angle>). This means a couple of things. We can see that the sine of 5π/7 is a positive number. In a similar manner, the sine of 9π/7 is a negative number. Further, since they are the same distance from π, they are the same quantity, except that one is positive and one is negative. In other words, by consideration of where the angles are, we can deduce that they in fact add to zero. We have thus confirmed that

sin(5π/7) + sin(9π/7) = 0

is a true statement! Thus, the equation as presented holds true.

By the way, the last term of the left half of the equation can be cos(6π/7), and the equality still holds true. Why?

Consider our diagram again. We know that cos(6π/7) and cos(8π/7) are also the same distance from π. We don’t know that value, but in the case of the cosine function we know both values are negative. Thus, whether we say:

cos(2π/7) +cos(4π/7) + cos(8π/7) = -1/2

or

cos(2π/7) +cos(4π/7) + cos(6π/7) = -1/2

we have a true statement, because cos(6π/7) = cos(8π/7).

]]>sin(5π/7) + sin(9π/7) = 0

is a true statement! Thus, the equation as presented holds true.

By the way, the last term of the left half of the equation can be cos(6π/7), and the equality still holds true. Why?

Consider our diagram again. We know that cos(6π/7) and cos(8π/7) are also the same distance from π. We don’t know that value, but in the case of the cosine function we know both values are negative. Thus, whether we say:

cos(2π/7) +cos(4π/7) + cos(8π/7) = -1/2

or

cos(2π/7) +cos(4π/7) + cos(6π/7) = -1/2

we have a true statement, because cos(6π/7) = cos(8π/7).

This question relies on two things:

1. How observant you are

2. Knowledge of mathematical priority (or knowing what is supposed to be done first).

To start, there are seventeen 1's in the question. Summing all of them would give 17. But as can be seen on the end, the last 1 is multiplied by zero. So we can eliminate 17 as an answer.

So the question now is: do we do that 1x0 before all that addition, or last?

In math, the order of operators is as follows:

Always evaluate anything in parenthesis first

Then, do multiplication or division from left to right

Lastly, perform addition or subtraction from left to right

So, we need to evaluate the 1x0

Nope! This is where the problem tests how observant you are. Look at the eleventh 1 in the problem: it is actually

1,2,3,4,5,6,7,8,9,10,9,10,11,12,13,14,14

Either way, the answer is 14.

It's surprising that about half of all the answers are 0. In fact, only about a third of all responders got the answer correct! Hey... anyone need some math tutoring? :-)

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Consider what we are asking with that question. If we say, for instance,

We get that solution by multiplying both sides of the equation by zero:

If

If

Anyway, we wind up with an infinite number of answers, or we wind up with an impossible answer. Either way, it just doesn't work.

Some people maintain that anything over zero is infinity, reasoning that the smaller the denominator gets, the larger the fraction. For instance:

1 / 1 = 1

1 / 0.1 = 10

1 / 0.01 = 100

1 / 0.001 = 1,000

1 / 0.0001 = 10,000

and so on. The reasoning goes that by extrapolating that out, we would get

1 / 0 = ∞

And since this could apply for any non-zero numerator, it is then tempting to say that

x/0 = ∞

Seems reasonable enough. Problem is, we could then do some fuzzy math along the lines of

since 1 / 0 = ∞

and 2 / 0 = ∞

and ∞ = ∞

one could say

1 / 0 = 2 / 0

And, multiplying both sides by zero, we'd get

1 = 2

Which isn't true. It proves that our initial definition

x/0 = ∞

is a fallacy. It just doesn't work.

After all, think of the definition of division. It's taking a quantity and dividing it into equal parts. Can you imagine taking six apples and dividing them among

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