## Friday, February 14, 2014

### 28: Trigonometric Identity?

Express the value of s as a rational number in lowest terms where

\$ s = & sin^2(10^{\circ}) + sin^2(20^{\circ}) + sin^2(30^{\circ}) +\\ &sin^2(40^{\circ}) + sin^2(50^{\circ}) + sin^2(60^{\circ}) +\\ &sin^2(70^{\circ}) + sin^2(80^{\circ}) + sin^2(90^{\circ}) \$?

And NO Calculator allowed.
UVM 2004-20

1. That's a fun question. It reminds me of this one involving logarithms:

http://i.imgur.com/oa92GO4.png

[Sorry, I do not know how to type LaTeX into a blogspot comment.]

1. I coach my high school's math team, and we just had a competition last Saturday. One of the parts on the "team" problem was similar to the one you link to with the jpg, though that one went all the way to log_31_(32).

2. Use \$ to begin it and end it. _5 is subscript 5. ^ for exponent. {} are the grouping symbols.
log_2 3 * log_3 4 * log_4 5 *log_5 6 * log_6 7 * log_7 8

3. \$log_2 3 * log_3 4 * log_4 5 *log_5 6 * log_6 7 * log_7 8\$

4. I've made this the Day 52 entry. Thanks for sharing!

5. As I mentioned in my reply to Anonymous, I coach my school's math team, so this is nothing new for me. The first competition of the year, the Trigonometry event covers identities, so this is one that's been asked before.

Note that since sin(x) = cos(90° – x), we know sin²(10°) = cos²(80°), sin²(20°) = cos²(70°), sin²(30°) = cos²(60°), and sin²(40°) = cos²(50°).

Thus, we can substitute and write s = sin²(10°) + sin²(20°) + sin²(30°) +sin²(40°) + cos²(40°) + cos²(30°) + cos²(20°) + cos²(10°) + sin²(90°).

Since sin²(y) + cos²(y) = 1, we can group the sin/cos pairs for 10°, 20°, 30°, and 40°, and get s = 4 + sin²(90°). Since sin(90°) = 1, and 1² = 1, we have s = 5.

* * * * *

The thing that's always surprised me about this problem is that the kids don't "see" it that quickly. For me, one of the things I first realized when I started learning basic right triangle trig is the relationships between the trig values for 30° and 60°. For me, I never think about 60° (or π/3) when I solve trig problems—I always convert to 30° (or π/6). I always figured it was one less thing to worry about memorizing. So with that understanding at hand, when I first saw this problem as a 16-year-old, it didn't give me any problems. I instinctively expected that the 10/80s were related, the 20/70s, etc. I confirmed that with the formula, and ran with it.

Since that came so naturally to me, when I teach Trig, I try to point it out to them as well. But they don't seem to gravitate to this really neat trig at all. This is when they give me that look and say, "We're not going to be math majors like you!" I tell them, "Hey, I'm just trying to make less memorization for you." Cause they think memorizing things is the worst thing ever. :-p