This article includes the following sections:

- Ways to Challenge Advanced Students
- How to Make Challenging Algebra Problems
- A Couple of Sample Problems (from my latest book)
- 50 Challenging Algebra Problems – Fully Solved (my latest workbook)

Some students catch on quickly, while others struggle to understand.

So how can you engage and challenge the better problem-solvers without blowing the minds of the other students?

- If you’re a teacher, you can include a challenging bonus question on an assignment, quiz, or exam.
- If you’re a parent, you can search for supplemental material with challenging problems.
- If you’re a student, you can try solving harder unassigned problems, reading ahead in your textbook, or looking for supplemental material.

For teachers (or very knowledgeable parents) who would like to make challenging problems, here are some ideas for how to go about it:

- Disguise the quadratic equation. One way is to make more than the usual three terms and put some on both sides of the equation, so that students need to combine like terms. Another way is to substitute y = x
^{c}, where c is a power of your choice. For example, if you let c = 2, your equation will look like a quartic, but if students make the substitution y = x^{2}, the quartic will turn into a quadratic. Another example is to let c = 1/2 and replace y = x^{1/2}with a squareroot. When students see that squareroot, it won’t seem like a quadratic.**Tip**: Think of what you want the answers to be, like x = 2 and x = 4/3, and then f.o.i.l. this into the quadratic, like (x – 2)(x – 4/3) = 0. (In this example, I would multiply both sides by 3 to get rid of the fraction.) - Make a system of equations look different from normal. Students are used to seeing things like 3x + 2y = 21 and 4x – 5y = 5. Instead, you might write the second equation like xy = 15, or y = 15/x. Another variation is to replace x with 1/x and y with 1/y. In my example, you would get 3/x + 2/y = 21 and 4/x – 5/y = 5. If students define t = 1/x and u = 1/y, they could then solve for t and u like usual, and then solve for x and y from t and u.
**Tip**: When making a system, first decide on the answers, like x = 9 and y = 12, next make the coefficients, and lastly determine the constants. - Put squareroots in problems they should know how to solve, but which usually don’t have squareroots in them. For example, if you have squareroot(3) and 1/squareroot(3) in the same equation, the terms can be combined by rationalizing the denominator to rewrite 1/squareroot(3) as squareroot(3) / 3. (Multiply the numerator and denominator by the squareroot of 3, and use the rule that the squareroot of 3 times itself equals 3.) Then factor out squareroot(3) to combine the terms.
- Put variables inside of squareroots. For example, you could write something like squareroot(x + 8) = x + 2. When you square both sides, you f.o.i.l. out the (x + 2) squared. There are many other ways to write a solvable equation with a variable in a squareroot. You could have things like squareroot(x) or 1/squareroot(x).
- Include an extra variable that cancels out. If students count equations and unknowns, it will seem like the problem can’t be solved. They’re partly right: The variable that cancels out will be indeterminate. But since one variable cancels, in this case it will be possible to solve for the other unknown(s).
- Use fractional exponents like 2/3 or 3/4. For example, if you isolate x to the power of 2/3, you can solve for x by raising both sides of the equation to the power of 3/2. You can make the numbers work out so that a calculator isn’t needed for students who understand the fractional powers. For example, if you raise 8 to the power of 2/3, you can do this in your head, since the cube root of 8 equals 2 (the power of 1/3 means to take the cube root, and you can think of 2/3 as involving a square and a cube root in any order). Of course, you can easily check your answer with a calculator, just to be sure.
- Put the unknown in a denominator. It’s amazing how many students freeze up over this, yet it’s not uncommon in science for variables to appear in the denominators of formulas. The progression of this occurs when students need to make a common algebraic denominator to solve for the unknown.
- Give problems that involve inequalities, especially when there are minus signs. For example, make a system of inequalities that involves substitution.
- Look for applications of algebra in higher-level math, physics, engineering, chemistry, economics, astronomy, and other subjects. (Sometimes, the problem isn’t really that hard, but it just seems hard because it’s out of context. But there are also challenging applied algebra problems in many subjects.)
- Make word problems that apply algebra. There is great variety in what can be done in the way of algebraic word problems.
- Prepare an elaborate solution that has an intentional mistake and ask students to figure out what is wrong. Ideally, the mistake would seem subtle, and would be a common mistake that many students tend to make on their own (like not distributing a minus sign correctly).

There are two types of problems that especially appeal to me:

- One kind where the problems look harder than they really are. I love it when students feel convinced that the problem is impossible based on what they know, but when they eventually figure it out on their own. This happens sometimes when a problem appears out of context, involves an application of algebra, or when a simple substitution makes something seem a lot more complicated than it really is. Word problems often fall in this category, too.
- The other kind is where the problem looks much easier than it really is. The structure might be so simple that at first glance you expect it to be a piece of cake, but there is some subtle aspect to it that it turns out to be much harder than it seemed. The catch is that the problem still needs to be reasonably solvable, so that the top students can still figure it out given enough time and sufficient background. It’s surprising how many mathematical problems look easy, but turn out not to be nearly as easy as they look.

Here are a couple of sample problems from my latest math workbook.

My latest math workbook includes 50 challenging algebra problems, space to try and solve them (in the paperback edition), and full solutions on the page following each problem (you won’t be able to see the full solutions until you turn the page).

There is a healthy variety of problems, involving a range of techniques. There is also a good range in the level of difficulty from problem to problem.

50 Challenging Algebra Problems (Fully Solved)

Paperback ISBN: 19416912344

Kindle ASIN: B07C7WHV3R

Copyright © 2018 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

**Newest releases** (in math):

- 50 Challenging Algebra Problems (Fully Solved)
- Fractions Essentials Workbook with Answers
- 300+ Mathematical Pattern Puzzles

**Improve Your Math Fluency**. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing
- physics

Not everyone loves fractions, but fraction skills are important for a number of reasons.

- It helps to be fluent with fraction skills when you take algebra, trigonometry, and calculus courses. When you solve for an unknown in algebra, the answer doesn’t always turn out to be a whole number. One reason that cross multiplication is difficult for many students is that the problems inherently involve fractions. In calculus, simple polynomial anti-derivatives naturally involve fractions (like the integral of x squared, which is one-third of x cubed plus a constant).
- We also use fractions in science courses like chemistry and physics. So if a student is struggling with fractions in math courses, these struggles also impact science courses. On the other hand, if a student takes the time to master fractions in math, this becomes an asset when fractions appear in other courses. Decimals and percentages (which are basically just other forms of fractions where the denominator is a power of ten) are very common forms of fractions in science and engineering.
- Fractions are actually pretty common. Almost every day, we see something happen a few times, and start to wonder how often that happens. Maybe your cell phone is doing something funny, or maybe you notice a quirk in somebody’s personality. If you give attention to this, you’ll realize that it’s common to wonder, “How often does that happen?” Many people just guess at it, but their guesses aren’t always realistic (especially when something obviously isn’t extremely common, but they say something like 90% of the time). If you have a small sample, like it happened 3 out of the last 8 times, you can use the fraction (in this case, 3/8) as a projection. You might convert 3/8 to a percentage and see that it equates to 37.5%, for example.
- If you have a ruler marked in inches, you will see fractions in many common measurements. The fraction will be something like 5/12, 7/8, 3/4, or 1/2. If you have a good feel for fractions, it helps to interpret these numbers. For example, if you measure the diameters of two different balls, and one ball is 3/8″ while the other is 1/3″, can you tell which is bigger just from the numbers?

With the importance of fractions in mind, following is my list of essential fraction skills, based on my experience helping students learn how to apply their math skills in science classes and laboratories.

- Visual association. Students should be able to draw pie slices to represent fractions, or should be able to write down a fraction to represent a pie slice.
- Terminology. You can’t discuss fractions with anybody or understand a lecture about fractions if you don’t understand what the different words and phrases mean, like numerator, denominator, reciprocal, common denominator, reduced fraction, greatest common factor, decimal, percentage, proper fraction, improper fraction, mixed number, ratio, and proportion.
- Reducing fractions. Students should be fluent in reducing fractions down to their simplest form. For example, 8/12 reduces to 2/3 if you divide the numerator (8) and denominator (12) both by 4.
- Common denominators. Given two different fractions, like 4/7 and 3/5, students should be able to find a common denominator.
- Mixed numbers. Students should be able to convert between improper fractions and mixed numbers.
- Addition and subtraction. Students should be able to find a common denominator in order to add or subtract fractions.
- Multiplication. Students should be able to multiply fractions. (It’s easier than addition or subtraction.)
- Reciprocals. Students should know how to find the reciprocal of any fraction or whole number.
- Division. Students should know that dividing two fractions is equivalent to multiplying by the reciprocal of the second fraction.
- Decimals. Students should be able to convert fractions into decimals or decimals into fractions.
- Percentages. Students should be able to convert decimals into percentages or percentages into decimals.
- Repeating decimals. Students should be familiar with repeating decimals and how they relate to fractions.
- Word problems. You know you understand the concepts well and can apply them when you can solve a variety of word problems that involve fractions.

Copyright © 2018 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest releases (in math):

- Fractions Essentials Workbook with Answers
- 300+ Mathematical Pattern Puzzles

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing
- physics

What are powers of 10?

How do you multiply or divide powers of 10?

What are negative powers of 10?

What is 10 raised to the power of zero?

Compare positive and negative powers of 10. Note that the number of zeros is a little different with positive and negative powers of 10. See the chart below.

Examples:

Teachers (and home school educators or parents) are welcome to use the above material regarding powers of 10 for non-commercial educational purposes.

Copyright © 2017 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest release (in math):

- 300+ Mathematical Pattern Puzzles

Click here to visit my Amazon author page.

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing
- physics

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It’s interesting that algebra can be useful in arithmetic.

Since arithmetic is a prerequsite to learning algebra, most algebra courses don’t focus on how to apply algebra to do arithmetic.

However, you can apply your algebra skills to arithmetic problems.

Following is an example.

Suppose that you would like to multiply 58 times 58, but don’t have a calculator (or cell phone) handy.

One way to do this, which is often taught in elementary school, is basically to multiply 8 times 58, then multiply 50 times 58, and add these together.

Algebraically, what your elementary school teacher taught you is that 58 × 58 = (50 + 8) × 58 = 50 × 58 + 8 × 58.

You may not have realized it at the time, but your teacher was applying the distributive property of mathematics.

When you solve the problem this way, first you multiply 8 × 58 to get 464. Then you multiply 50 × 58 to get 2900. Finally, you add these together to get 3364.

Now let’s see how algebra can supply an alternative solution.

I’d prefer to work with nice round numbers. Hey, 60 is a round number close to 58. If I could work with a round number, like 60, and a small number, like 2, that would make the arithmetic much simpler to do without a calculator.

So let’s write 58 as 60 minus 2.

Then the problem is 58 × 58 = (60 – 2) × (60 – 2).

Use the f.o.i.l. method from algebra. Recall that the f.o.i.l. method stands for “first,” “outside,” “inside,” “last.”

One example of the f.o.i.l. method from algebra is (x + y)(x + y) = x² + xy + xy +y² = x² + 2xy +y².

Let’s apply this to the arithmetic problem. We can think of (60 – 2) as (x + y), where x = 60 and y = –2.

58 × 58 = (60 – 2) × (60 – 2) = 60 × 60 + 2 (60)(–2) + (–2) × (–2).

Look what we did. We wrote the original problem, 58 × 58, in terms of three simple multiplication problems: 60 × 60, 2 × 60 × (–2), and (–2) × (–2).

58 × 58 = 3600 – 240 + 4 = 3364.

Note that (–2) × (–2) = +4 because the two minus signs cancel out.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest release:

- 300+ Mathematical Pattern Puzzles

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing

Here is an exercise in pattern recognition.

It’s not a linear pattern.

This is an array, so there is a slight geometric element to the pattern.

See if you can figure out the missing numbers in the above puzzle.

Study the four arrays.

See if you can recognize the pattern.

Once you identify the pattern, apply it to the fifth array.

Spoiler alert.

If you scroll down too far…

You will run into the answer.

So stop scrolling down…

If you would like more time to solve the puzzle.

Ready or not.

Here it comes.

First the answer:

Now the solution.

Begin with the top left number.

Double the top left number. That makes the top right number. 5 doubled = 10.

Now multiply the top two numbers. That makes the bottom left number. 5 times 10 = 50.

Now add the bottom left number to the top right number. That makes the bottom right number. 50 plus 10 equals 60.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest releases:

- 300+ Mathematical Pattern Puzzles
- Basic Linear Graphing Skills Practice Workbook
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing

Here is a math puzzle challenge.

**Hint**: Each of the 5 patterns below has something in common.

**Directions**: See if you can figure out which numbers go in the blanks.

- 1, 2, 4, 6, 10, 12, 16, 18, 22, _, _
- 4, 6, 10, 14, 22, 26, 34, 38, _, _
- 3, 7, 13, 19, 29, 37, _, _
- 4, 9, 25, 49, 121, 169, 289, _, _
- 5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, _, _

If you need help, you can find hints below.

But don’t scroll too far or you’ll run into the answers and explanations.

Each pattern above has something in common.

They all involve prime numbers.

A prime number is only evenly divisible by two integers: 1 and itself.

For example, 7 is a prime number because the only integers that can multiply together to make 7 are 1 and 7.

In contrast, 6 isn’t a prime number because 2 x 3 = 6 (in addition to 1 x 6).

Here are the first several prime numbers.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Each of the puzzles above relates to these prime numbers.

When you’re ready, you can find answers and explanations below.

Here are the answers and explanations to the math puzzles:

- 28, 30. Explanation: Subtract 1 from each prime number: 2 – 1 = 1, 3 – 1 = 2, 5 – 1 = 4, 7 – 1 = 6, 11 – 1 = 10, etc.
- 46, 58. Explanation: Double each prime number: 2 x 2 = 4, 3 x 2 = 6, 5 x 2 = 10, 7 x 2 = 14, 11 x 2 = 22, etc.
- 43, 53. Explanation: Every other prime number: 3 (skip 5) 7 (skip 11) 13 (skip 17) 19 (skip 23) 29 etc.
- 361, 529. Explanation: Square each prime number: 2² = 4, 3² = 9, 5² = 25, 7² = 49, 11² = 121, etc.
- 78, 84. Explanation: Add consecutive prime numbers together: 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, etc.

One way is to follow my blog. I will post occasional math puzzles in the future.

Another way is to check out my newest book, 300+ Mathematical Pattern Puzzles.

It starts out easy and the level of challenge grows progressively so that puzzlers of all abilities can find many puzzles to enjoy.

A wide variety of topics are covered, including:

- visual patterns
- arithmetic
- repeating patterns
- Roman numerals
- Fibonacci sequence
- prime numbers
- arrays
- analogies
- and much more

The cover was designed by Melissa Stevens at www.theillustratedauthor.net.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest releases:

- 300+ Mathematical Pattern Puzzles
- Basic Linear Graphing Skills Practice Workbook
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing

Slope is a measure that indicates how steep or shallow a straight line is:

- A line with greater slope is steeper.
- A line with less slope is shallower.
- A horizontal line has zero slope.

Slope can be positive or negative:

- A line with positive slope slants upward.
- A line with negative slope slants downward.
- A line with zero slope is horizontal.

The slope of a straight line equals rise over run.

- The rise between two points is vertical. It’s the change in y.
- The run between two points is horizontal. It’s the change in x.

From the graph of a straight line, determine the slope as follows:

- Mark two points on the line.
- Read the x- and y-coordinates of the two points, (x1, y1) and (x2, y2).
- Subtract y2 – y1 to get the rise.
- Subtract x2 – x1 to get the run.
- Divide the rise by the run.

Here are a few tips:

- When choosing the two points, try to find points where both x and y are easy to read without interpolating. This isn’t always possible: In that case, at least one coordinate should be easy to read without interpolating.
- Choose two points far apart. This reduces the relative error in interpolating.
- Make sure that both points lie on the straight line. Don’t choose a point that’s close to the line, but doesn’t lie on it.

**Example**: Find the slope of the straight line in the graph below.

**Solution**: First, choose two points on the line. Ideally, these points should be far apart and easy to read.

In this case, it’s easy to read both the x- and y-coordinates for the leftmost and rightmost points shown in the graph. So let’s choose those.

- The leftmost point has coordinates (0, 3).
- The rightmost point has coordinates (10, 8).

Subtract the y-values to determine the rise:

y2 – y1 = 8 – 3 = 5

Subtract the x-values to determine the run:

x2 – x1 = 10 – 0 = 10

(In coordinate graphing, recall that x is horizontal and y is vertical.)

Divide the rise by the run to find the slope:

The slope of the line is 0.5.

**Check**: You can check your answer as follows.

Look at the graph. Starting from (0, 3), the next point that’s easy to read is (2, 1).

From (0, 3) to (2, 1), the line goes one unit up (vertically) and 2 units over (horizontally).

The ratio of the rise to the run is 1 to 2. Divide the rise (of 1) by the run (of 2). The slope is 0.5.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest release:

- Basic Linear
**Graphing Skills**Practice Workbook

Related books:

- Trigonometry Essentials Practice Workbook with Answers
- Learn or Review Trigonometry Essential Skills
- Algebra Essentials Practice Workbook with Answers
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

There is a simple way to remember the sine, cosine, and tangent of special trigonometry angles.

The special trig angles are 0º, 30º, 45º, 60º, and 90º. What makes these angles special? The 30º-60º-90º triangle is one-half of an equilateral triangle, while the 45º-45º-90º triangle is one-half of a square. In both cases, the trig functions (sine, cosine, and tangent) can be expressed as simple ratios.

Here is the trick for quickly working out the sine, cosine, and tangent of 0º, 30º, 45º, 60º, and 90º.

Write the special angles in order.

Write the integers 0 thru 4 in order.

Squareroot each number.

Divide each number by 2.

These are the sine of 0º, 30º, 45º, 60º, and 90º.

It’s that simple. Here’s a recap:

- Write the numbers 0 thru 4.
- Squareroot each number.
- Divide each number by 2.

Just write the previous numbers in reverse order.

Why does this work? Because the sine of theta equals the cosine of the complement of theta: sin(*θ*)=cos(90º–*θ*). What’s opposite to theta is adjacent to its complement.

Divide sine theta by cosine theta.

This chart shows all of the steps together.

- Write the special angles.
- Write the integers 0 thru 4.
- Squareroot each number.
- Divide each number by 2. This gives you sine of theta.
- Write the numbers in reverse order. This gives you cosine of theta.
- Divide the previous two rows (sine over cosine). This gives you tangent theta.

There are two different, yet equivalent ways, to write the above chart.

That’s because of the following properties of irrational numbers:

So, for example, there are alternative ways to express the following trig values:

The chart on this blog uses **standard form**. Most math courses use standard form, which means that there no irrational numbers (like root 2) in the denominator.

Sometimes you find the trig table in another nonstandard form. In that form, you see 1 over root 2 in place of root 2 over 2, and you see 1 over root 3 in place of root 3 over 3.

It’s important to realize that both forms are correct. The standard form, however, is expected in most math courses.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

- Trigonometry Essentials Practice Workbook with Answers
- Learn or Review Trigonometry Essential Skills
- Trigonometry Flash Cards (for Kindle)
- Algebra Essentials Practice Workbook with Answers
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule
- Other volumes cover fractions, long division, arithmetic, and more
- Also look for books on the fourth dimension, astronomy, conceptual chemistry, and more

When students learn multiplication, it starts out easy.

Too easy, as 0 times anything equals 0 (e.g. 0 x 8 = 0, while anything times 1 equals itself (e.g. 4 x 1 = 4).

But it doesn’t stay easy. Most students struggle to remember multiplication facts where one of the numbers is 6, 7, 8, or 9.

However, there are some tips to aid in memorization.

For one, mirror images don’t matter. The word for this is commutative.

This means 2 x 8 = 8 x 2, for example. Both equal 16.

So if you know 6 x 8 = 48, you also know that 8 x 6 = 48. You don’t need to memorize both.

One of the tricky multiplication facts is 7 x 8. But it’s easy if you remember the trick.

Remember 5678. These are 5 thru 8 in order. 56 = 7 x 8. Piece of cake, huh?

The 9’s are easy. The answer is one decade less than multiplying by 10, then make the two digits add up to 9.

For example, 9 x 7 is in the 60’s (because 10 x 7 = 70, one decade less is 60). It’s 63 since 6 + 3 = 9.

Another example is 5 x 9. It’s in the 40’s (one decade less than 5 x 10 = 50). It’s 45 since 4 + 5 = 9.

To do the trick, after you figure out the decade (by multiplying by 10 and then subtracting 10), subtract the tens digit from 9 to get the units digit.

For example, consider 9 x 6. Multiply 10 x 6 to get 60, and subtract 10 to make 50 (one decade less). Now subtract 5 (the tens digit of 50) from 9 to get 4. Therefore, 9 x 6 = 54.

Once you get the hang of it, this makes remembering the 9’s easy. Try out all the 9’s to get some practice.

If you’re good at doubling numbers quickly, try writing 6 as 2 x 3.

Then 6 x 7 = 2 x 3 x 7. If you know 3 x 7 = 21, double 21 to get 42.

Similarly, for 6 x 4 = 2 x 3 x 4, start with 3 x 4 = 12 and double 12 to make 24.

You can use the doubling trick for the 8’s, too. Just double the number 3 times.

For example, consider 5 x 8. Double 5 three times: 10, 20, 40. So 5 x 8 = 40.

Try 8 x 6. Double 6 three times: 12, 24, 48. Therefore, 8 x 6 = 48.

It works with the 4’s, also. Just double twice.

With 4 x 9, double 9 twice: 18, 36. So 4 x 9 = 36.

That leaves 7 x 7 = 49. You should know 7 x 9 from the 9’s trick.

You can make 7 x 6 and 7 x 8 from the doubling tricks. (The latter you can also know from the 5678 trick.)

5 and under are easier. So to complete the 7’s, you really just need to memorize 7 x 7 = 49.

This covers the 6 thru 9’s, which tend to be the trickier multiplication facts.

The 10’s and 11’s are easy. For the 10’s, just add a zero, as in 8 x 10 = 80 or 6 x 10 = 60.

For the 11’s, with 1 thru 9 just double the digits, like 3 x 11 =33 or 11 x 8 = 88. Get 11 x 10 = 110 from the 10’s trick (add a zero). Then you just need to memorize that 11 x 11 = 121 to complete the 11’s.

You can get the 12’s by doubling the 6’s. For example, knowing that 6 x 5 = 30, you can find that 12 x 5 = 60 by doubling 6 x 5.

Do you know any other tips for remembering multiplication facts 0-12? If so, please share them in the comments.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

- arithmetic facts
- multi-digit arithmetic
- long division with remainders
- fractions, decimals, and percents
- algebra and trigonometry