Solution to The October 15 , 2006 PoW

I have read in some math manuals that a quick way to check whether your calculator is in radian mode or degree mode is to take the Cosine of 90. If your calculator is in degree mode the answer is 0, and not 0 (~ -0.448) if your calculator is in radian mode. Now if we knew nothing about trigonometry we might prefer to take the Cosine of 0. But we do know something about trigonometry. The Cosine of 0 is 1 (cos(0) = 1) in both degree and radian mode, so 0 cannot be used to check the mode of your calculator. My question to you is:

Find all numbers so that the Cosine of that number is the same in degree mode as it is in radian mode. In fact, we might as well also do this for the Sine and Tangent functions.
Exact answers only, but here is a decimal approximation of one value that works for sine:

x ~ 3.08770208587 implies sin(3.08770208587) ~ 0.05386448674 in both radian and degree mode.

Solution to this trigonometry problem. Use trig identities.

First, suppose we want to find all x's so that tan(x) gives the same value in both radian and degree mode. If our calculator is in radian mode then we need to solve . That is, we want to find all x's so that the tangent of the angle in radians equals the tangent of the angle in degrees. Since our calculator is in radian mode, we convert the angle in degrees to radians (this tells us what the tangent of x degrees is). Therefore, or

sin(x)cos(xp/180) - cos(x)sin(xp/180) = 0. Using the identity sin(A – B) = sin(A)cos(B) – cos(A)sin(B), the above equation becomes sin(x-xp/180) = 0. Since sin(t) = 0 if and only if t is a multiple of p, we must have x-xp/180 = np, where n is an integer. The solution set of all values so that tan(x) in radian mode equals tan(x) in degree mode is the set of all x ,, where n is any integer.

Now suppose our calculator is degree mode and we wish to find all x's so that cos(x) in radian mode equals cos(x) in degree mode. Let us suppose we have a calculator in degree mode. Then we want to find all x's so that cos(x) = cos(180x/p). (The latter converts x from radians to degrees, which we need to do to find the cos(x) radians when our calculator is in degree mode.)

So cos(x) - cos(180x/p) = 0, or sin((px+180x)/2p)sin((px-180)/2p) = 0. (Since Cos(A+B) - Cos(A-B) = -2Sin(A)Sin(B).) Therefore either (px+180x)/2p = 180n for some integer n, or (px-180)/2p = 180m for some integer m. So the solution is the set of all x of the form where n and m range over the integers.

We can use the trig identity Sin(A+B) - Sin(A-B) = 2Cos(A)Sin(B) to solve for all x's so that sin(x) = sin(xp/180) (assuming our calculator is in radian mode). But instead let's note that we can also look at the problem as finding all x so that sin(x + 2np) = sin(xp/180) for some integer n (since

sin(x) = sin(x + 2np) for any integer n if our calculator is in radian mode). So if (x + 2np) = (xp/180),
this gives where n ranges over the integers as solutions. This is only part of the total solution set. To get the other part, note that sin(x) = -sin(x+(2n-1)p). So to solve sin(x) = sin(xp/180) we can write -sin(x+(2n-1)p) = sin(xp/180). Or, sin(x+(2n-1)p) = sin(-xp/180) for some integer n.
Therefore where n ranges over the integers is also part of the solution set. That is, the final solution set of all x's so that sin(x) in radian mode equals sin(x) in degree mode is

where n and m range over the integers.