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Kara E.
Experienced High School/Middle School Math Teacher
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Trigonometry
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Question:

Verify that $$(\sin(x)-1)(\tan(x)+sec(x))=-\cos(x)$$ is an identity.

Kara E.

This problem might seem pretty scary to start out with, but I promise I won’t leave you hanging here- we are going to work through this problem, step by step, and by the end you’ll think “I was scared of that little thing?!” The most important part of verifying trig identities (or any proof for that matter) is don’t be scared! Just try it! The worst that can happen is you’ll get stuck, and have to backtrack a bit, but learning how NOT to do something is just as important as learning how TO DO something (isn’t there a quote by Benjamin Franklin about that?). I always think of verifying identities and proofs as kinda like a maze. You know where you’re starting, and you know where you need to go, it’s just the middle part that gets crazy, and sometimes when you’re doing a maze, you go down a path and realize you’re at a dead-end, so you backtrack, and try again. Sometimes you may hit a couple dead-ends, but that’s ok, as long as you persist and get to the end, even if you go the super long way around the maze, you finished it and that’s all that counts! So you decided you're going to start the problem. Now where do you start? When I taught this to my high schoolers, I would start out with a few tips: First, see if any part or parts of the original question can be substituted with an identity that you’ve already learned. One example is seeing $$cos^{2}x+sin^{2}x$$ which always equals 1, so do that first! Check to see if any parts of the problem stick out to you as familiar, and work with those. Second, if there are a bunch of tangents or cosecants or secants in there, change them all to variations of sine and cosine. Things will be a lot easier to simplify if we only have two different things to work with instead of 6. Lastly, try to work with one side of your identity and get as far as you can. Some students want to start changing both sides of the problem, which is a lot like drawing two lines in a maze- one at the start and one at the end, and trying to meet somewhere in the middle. Let me tell you, it gets pretty crazy when we do that. So try to work with one side at a time. Sometimes you get towards the end, and you’re so close, and it’s easier to change the side you left alone than to try and tweak the working side to finish the problem, and that’s ok. As long as you end with both sides of the equal sign showing the exact same thing, you did it! Now we know how to start, so let's do it! So this problem seems crazy with the sine and the tangent and adding/subtracting and then multiplying it all?! What?! It's ok, start with the tips I gave you. Are there any familiar parts to the problem? We see $$\sin(x)-1$$ which is close to an identity ($$\sin^{2}(x)+\cos^{2}(x)=1$$) but not quite, so we move on to the next tip. We have a couple different trig ratios that aren't fun to use in verifications, so let's change the tangent and secant to variations of sine and cosine. We will change $$\tan(x)$$ to $$\frac{\sin(x)}{\cos(x)}$$ and $$\sec(x)$$ to $$\frac{1}{\cos(x)}$$. Now our problem looks like this: $$(\sin(x)-1)(\frac{\sin(x)}{\cos(x)}+\frac{1}{\cos(x)})$$ Still looking pretty crazy, huh? Remember, things might get more messy before they start to get better. Just because it's getting crazy doesn't mean you made a mistake. Let's keep going. Remember the last tip- only work with one side of the equals sign. Since there's a billion different things we can do with $$-\cos(x)$$ but only a limited amount of ways we can simplify $$(\sin(x)-1)(\frac{\sin(x)}{\cos(x)}+\frac{1}{\cos(x)})$$, let's stick with working on the left side for now. We have two sets of parenthesis that somehow need to combine to be one term, so let's multiply them, since we can't do much inside the parenthesis. **Side note: if you saw here that both terms inside the second set of parenthesis have a cosine in the denominator and wanted to combine those two terms to be $$\frac{\sin(x)+1}{\cos(x)}$$ that would also be a good next step. There are a TON of different ways to verify identities. The way I'm working through is only one option. So we decided we're multiplying. We have two binomials multiplied together, so we need to use the FOIL method (first, outside, inside, last), or whatever method you were taught/is easiest for you. Personally, I use the FOIL method, so I'm going to explain it that way. First terms are $$\sin(x)$$ and $$\frac{\sin(x)}{\cos(x)}$$ so we multiply these to be $$\frac{sin^{2}(x)}{\cos(x)}$$ (remember when you have a plain term like $$\sin(x)$$ that means that it's in the numerator position. 1 would be in the denominator). When we multiply the outside terms $$\sin(x)$$ and $$\frac{1}{\cos(x)}$$ we have $$\frac{\sin(x)}{\cos(x)}$$. Inside terms -1 and $$\frac{\sin(x)}{\cos(x)}$$ multiply to be $$-\frac{\sin(x)}{\cos(x)}$$. Finally, last terms -1 and $$\frac{1}{\cos(x)}$$ multiply to be $$-\frac{1}{\cos(x)}$$. Now our problem looks like this: $$\frac{\sin^{2}(x)}{\cos(x)}+\frac{\sin(x)}{\cos(x)}-\frac{\sin(x)}{\cos(x)}-\frac{1}{\cos(x)}$$ See how those middle terms are opposites of each other? They're the exact same thing but one is positive and one is negative! Let's simplify that to be 0. Now our left side is looking far less complicated. It looks like this: $$\frac{\sin^{2}(x)}{\cos(x)}-\frac{1}{\cos(x)}$$ OK, we are making great progress! See how we have two fractions with the same denominator? We can combine these two fractions easily (remember adding and subtracting fractions in middle school? I hope so, because you'll use it a ton while doing trigonometric identities!) After we combine those, we have this: $$\frac{\sin^{2}(x)-1}{\cos(x)}$$ Our numerator is looking pretty familiar. Remember that identity we tried to use at the beginning? $$\sin^{2}(x)+\cos^{2}(x)=1$$? It looks like we might be able to use it now! If we tweak it a little bit, we can get the $$\sin^{2}(x)$$ and the 1 on the same side, and we have $$\sin^{2}(x)-1=-\cos^{2}(x)$$ After we plug that in for the numerator, we have $$\frac{-\cos^{2}(x)}{\cos(x)}$$ We are so close to the end! We have 2 cosines on the top, and one on the bottom, so we can simplify this to 1 cosine on the top, and 0 cosines on the bottom, which would be $$\frac{-\cos(x)}{1}$$ (Many people make the common mistake of simplifying fractions and putting a 0 in the denominator. A 0 on the bottom of a fraction is a very dangerous offense! Watch out for that because that might be a red flag that something went wrong in your proof. You should NEVER divide by 0) Now we have this: $$-\cos(x)$$ Wait a minute, that's what the right side is! That means we're done! It took a little bit, and things got messy there for a while, but we persisted and got to the end of our little math maze. And we were scared of that problem in the beginning. Ha!

Geometry
TutorMe
Question:

Explain how to find all the missing angle measures in the quadrilateral and the missing arc measures of the circle. Draw a circle. Draw 4 dots of varying distances around the edge of the circle. Label these dots A, B, C, and D. Now draw lines connecting these four dots to make a quadrilateral (does not have to be a square). Now label the following things: Angle ADC is 94 degrees Arc DC is 70 degrees Arc BC is 78 degrees. We need to find the other three angle measures of the quadrilateral (DAC, ABC, and BCD) and the missing two arc measures of the circle (AB and AD)

Kara E.

Algebra
TutorMe
Question:

Fully explain how to factor the quadratic polynomial $$2x^{2} + 7x + 3$$.

Kara E.

The first thing I do when I'm about to factor something is to check to see if there are any "shortcuts" I can use. The first shortcut I always look for is "Is there a number and/or variable that I can factor out of all the terms?" In this case, I would especially look for factoring out a 2 from all terms so the leading coefficient (the number at the front of the first term when the polynomial is written in standard form-- see definition of standard form below) is 1. This makes the polynomial MUCH easier to factor. Unfortunately, we cannot factor a 2 out of all three terms because 2 cannot be factored out of 7 or 3. The second shortcut I look for is "Is this a perfect square trinomial or the difference of two perfect squares?" (see definition of both of these terms below). For each of these, we have three questions to ask ourselves to check to see if these shortcuts will work. For the perfect square trinomial, we ask ourselves the following questions: 1) is the first term a perfect square? (see definition of perfect square below) 2) is the last term a perfect square? 3) is the middle term equal to 2*(the square root of the first term)*(the square root of the last term)? The answer for all three questions needs to be yes for us to use this shortcut. That is not the case here because 3 is not a perfect square (if we take the $$\sqrt{3}$$ we would get a decimal). For the difference of two perfect squares, we need a polynomial with only two terms. This polynomial has three terms, so we know it cannot be the difference of two perfect squares. If we did have two terms, we would ask ourselves the following questions: 1) is the first term a perfect square? 2) is the second term a perfect square? 3) is the sign between them a subtraction sign? Again, we cannot use this shortcut because our polynomial has three terms. So, we have to use the long hand way of factoring the polynomial. Let's go! Many teachers may teach factoring when the leading coefficient is not 1 in a variety of ways, but the method that I have found most successful for a variety of students is the following: Step 1) Multiply the coefficient from the first term with the last number. In this case we have $$2*3$$ which is 6. Now, we need to think of a pair of numbers that multiply to be 6, and add to be the coefficient of the middle term, which is 7 in this case. Two numbers that fit these two requirements are 1 and 6. We will use these numbers in step 2! Step 2) Split the middle term (7x) into 1x and 6x. Now the polynomial looks like this: $$2x^{2} + 1x + 6x + 3$$ It does not matter whether we put 6x + 1x in the middle or switch them around to be 1x+6x. Our answer will come out the same in the end either way. Once we get better at this, we can arrange the numbers to make our job easier, but for now, we will just put them in this way. Step 3) Factor the polynomial 2 by 2. What comes out of the first two terms? We can see the first two terms have an x in common, but no numbers, so we factor out an x. Now our polynomial looks like this: $$x(2x+1) + 6x+3$$ Now, do the same for the second set of numbers. What comes out of the second pair of numbers (6x and 3)? We can see that a 3 can be factored out of both numbers, so we do so. Now our polynomial looks like this: $$x(2x+1) +3(x+1)$$ Step 4) Now look at the whole picture. See how can factor a (2x+1) out of each section of the polynomial, the first section and the second section? Let's do that. We are almost done! Final step is to factor out the common terms (2x+1) from each part, and take the remaining parts of our polynomial (x and +3) and put them in a set of parenthesis right next to (2x+1). We are left with our answer: $$(x+3)(2x+1)$$ Again, it doesn't matter which order we put these two factors in, since multiplication (the invisible sign between the two sets of parenthesis) is commutative (meaning we can switch the numbers around and get the same result). Another way you could write the answer would be $$(2x+1)(x+3)$$ Now don't forget the most important part of any math problem-- checking your work! In the case of factoring, it is always very easy to check your work, all you have to do is multiply the two binomials inside the parenthesis (Using the FOIL method, if that's what you were taught). Your answer after you multiply them will be the same polynomial we started with $$2x^{2}+7x+3$$ and it is! Great work! Standard form: when a polynomial is written in decreasing degree. Ie, for quadratic polynomials, the $$x^{2}$$s come first, then the xs, then the constants. Perfect Square Trinomial: a polynomial that can be factored to be $$(ax+b)^{2}$$ or $$(ax-b)^{2}$$. Difference of Two Perfect Squares: a polynomial that can be factored into $$(ax+b)(ax-b)$$. ALWAYS in the form of a perfect square minus a perfect square. Needs to be a subtraction sign between the two perfect squares. If it is an addition sign, the polynomial cannot be factored in the form $$(ax+b)(ax-b)$$. Perfect Square: When you can take the square root of the number and/or variable, and have a whole number and/or variable left. Decimals are not easy to use in factoring, so we stay away from them at all costs.

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