#trigonomety

I’ve been very tired today. I struggled to get through this assignment, especially that last question when it’s actually simple if you use similarity ratios, since both measurements are taken at 3pm and then once having the height of the tree, you can find the angle. I didn’t write the full question. It was long. Both measurements are at 3pm; otherwise, similarity ratios would not be accurate since the shadow cast in one wouldn’t be similar with the other since the sun would be hitting the tree from a different angle.

Mom was here tonight but working online, so I watched the littles again. The best part is getting extra Hearthstone time this weekend and money added to my account balance so I can spend it on stuff. Sometimes my mom is cool.

Accomplished today:

Geometry assignment

English vocab test (100% after realizing I skipped 4 questions)

Finished Emma chapter 26

Comp assignment

Spanish assignment

Duolingo

WH assignment

Gave speech (100%)

Dance

Elective study assignment/journal

Mathematics

On April 6th, my 3rd grade math stayed the same at 99%, my 4th grade math improved from 85% to 92% and my trigonometry improved from 81% to 85%. 

Russian

I wrote probably the longest journal entry ever in Russian! I talked about my week and my plans for the weekend.

Quantum Computing

Solve for A and B.

Good move substituting the numbers with X and y. It's easier to see a formula. 

The corresponding formula is cos(x-y) = cosXcosY+sinXsinY.

So then we can just substitute back in, and we get cos((3pi/7)-(2pi/21)). Just add the insides, and we get cos(7pi/21). Reduce this down to cos(pi/3).

So then cos(pi/3) = cos(pi/A). This is a bit... eh.. because I'm not sure if this is a valid math move but you say that A = 3, because it's kind of simple. But I guess if you had to do something more math-y, take the inverse cosine of each side and then cross multiply and solve for A. 

So A = 3.

This is one of my favorite kinds of problems!! #nerd

Start by drawing a line down the page from the equal sign, separating the two sides of the equation. The left side is messier in this case, so we are going to manipulate it until we get  the right side, but we aren't going to touch the right side.

FOIL the binomial, so we get tan^2 (u) + 2tan(u) sec(u) + sec^2 (u).

Next, let's change everything into sin and cos, using identities. Remember that tan = sin/cos, and sec = 1/cos. So we get sin^2 (u) / cos^2 (u) + 2 (sin(u) / cos(u) )(1/cos (u) )+ 1/cos^2(u). 

Now, add everything up since htey all have the same denominator to get [sin^2 (u) + 2sin(u) + 1] / cos^2 (u). The top is a perfect square of a binomial, which factors into (sin(u) + 1 )^2, which = (sin(u) + 1)(sin(u) + 1). 

Looking at the bottom, we replace cos^2 (u) with 1-sin^2 (u) (because sin^2 + cos^2 = 1), and then that is a difference of squares, which factors into (1 + sin(u) )(1- sin(u) ).

So putting everything together right now, we have: [(sin(u)+1 )(sin(u)+1 ) ] / [(1+sin(u)) ( 1 - sin(u) ). The (1+sin(u)) cancel out, and we are left with:

(1+ sin u) / (1-sin u), which is what we have on the right side. so we are done!

These problems are challenging for people because they require a lot of implicit knowledge and being able to see what is hidden in there (the identities ) and a lot of knowledge of algebra skills.