Review: Trigonometic Functions

Trigonometic Funtions Part 1



Radian Measures

• 2radians = 360 degrees, radians = 180 degrees
• to convert degrees to radians, multiply degree by (/180)
• to convert radians to degrees, multiply radians by (180/)

Special Angles

• you can use the unit circle to solve
• use special angles with unit circle

Equivalent Trigonometric Expressions

• use the cofunction identity sinx = cos (/2 - x) to determine equivalent trigonometric expressions

Compond Angle Formula

• apply equivalent trigonometric expressions to compund angle formula
• determine exact trigonometric ratios for angles expressed as sums or differences or special angles

Identities

• treat each side independantly and transform expression on one side into exact form of the other

Trigonometric Functions Part 2

Sine, Cosine, and Tan Graphs

• y=sinx and y=cosx graphs have amplitude of 1 and a period of 2 and can be transformedy=tanx has no amplitude and no maximum or minimum

Reciprocals

• reciprocals are different from it's inverse
• csc means 1/sinx
• sec means 1/cosx
• cot means 1/tanx

Mapping Rule

• g(x) = af [k(x-d)] + c
• a represents amplitude
• d represents phase shift
• c respresents vertical translation

Period

• 2/k

Amplitude

• max - min / 2

6.1 The Exponential Function and its Inverse

This is our second last unit of our Advanced Functions course. YAYYYY ^^
In the lesson, it will be separated into two parts. The first part will talk about the exponential functions and the second part will talk about its inverse.

Since this is the first lesson of the new logarithmic unit, most of the information presented should be learned by students.

PART 1- EXPONENTIAL FUNCTIONS

In part 1, we will learn how to solve for the exponent with different bases. The "log" function on your calculators will be used greatly in this unit.

Side note: the default base of the "log" key on your calculators is ALWAYS 10.

PART 2- Inverse

In part 2, we will be pretty much doing the same thing as grade 11. Switch x and y in the same equation and isolate for X again.


Exponential functions can be represented in 3 DIFFERENT ways.

1: General Form- the simplest form of representation of an exponential equation

y=5^x or y=1/5^x

2. Table of Values- a table which a list of the "x" values with the corresponding "y" values

REMEMBER: the first ratio of differences of the x and y values are always constant.

How do we find the first ratio of differences?

The values of : Y₂-Y₁ = answer.

3: Exponential functions- this is a graph representation of the exponential functions. However, just by looking at a graph cannot help you determine whether an exponential function is or not.

Rate of Change:

To determine the rage of change of an exponential function is really much similar to finding the rate of change of any other functions.

1) Pick the x values that are really close to the value you are given to find the IROC.

2) For example: if the value that is given is 2. You may go to the left of the value or right to the value which results in x = 2.0001 or x = 1.9999.

3) Sub these values back into the original equation and find the “y” values

4) To choose the two closet points that lead to the given value will give you the best and most accurate slope of secant that is required for the answer.

5) After having the “x” values and the “y” values. Put it into the general equation used to find slopes which is:

AROC [x value – x value] = y₂-y₁ / x₂-x₁

6) To conclude your answer, you must write, “Therefore, the IROC of the given value is approximately at “answer of AROC of the 2 x values”.

Inverse functions:

The graph of the inverse function of a logarithmic function is a reflection on the y= x function.

To find an inverse function of a logarithmic function, switch x and y from the equation and solve for x again.

By switching (x,y) -> (y,x) , these can be your new inverse values that can be used to apply transformations.

Differences and similarities between an inverse function and an exponential function

1) A exponential function will have an increasing slope while the inverse function will have an decreasing slope

2) A horizontal asymptote (y = a value) is present in an exponential function. A vertical asymptote (x = a value) is present in the inverse function

3) Depending on the transformations, the horizontal asymptote or the vertical asymptote is affected by the vertical or horizontal shifts.

7.2 Techniques for Solving Exponential Functions

Hi, this post covers 7.2 Techniques for Solving Exponential Functions.

- Radioactive Decay
In this chapter, we learned how to find the half-life of radioactive substances, which means we know how to find the half-life of pretty much anything that works in a similar fashion. The equation we are given is:

A(t) = A₀(1/2)^t/h

All we have to do now is substitute a corresponding value in each variable.
t = time, h = half-life, A(t) = mass of substance after a period of time, and A_{0 =} mass of substance at 0 time. If you have any 3 of the 4 variables, then it is possible to isolate and solve for the unknown variable. An example is shown on page 370 of our textbooks.

- Powers With Different Bases
For these types of questions, we can either solve by Applying Algebraic Reasoning or Using a Graphing Calculator.

Applying Algebraic Reasoning:
For this method, we take the log of both sides and use the power law of logarithms.

4^2x-1 = 3^x+2

log4^2x-1 = log3^x+2

(2x - 1)log4 = (x + 2)log3

2xlog4 – log4 = xlog3 + 2log3

2xlog4 – xlog3 = 2log3 + log4

x(2log4 – log3) = 2log3 + log4

x = 2log3 + log4 / 2log4 – log3

After that, just use a calculator if you want to determine the approx. value of the expression. In this case, the answer is approximately 2.14.

- Apply the Quadratic Formula
Lastly, we learned how to solve questions like this:

2^x – 2^-x = 4

In order to use the quadratic formula to solve for this type of question, we must multiply both sides by 2^x in order to generate a suitable equation:

2^x(2^x – 2^-x) = 2^x(4)

2^x(2^x) - 2^x(2^-x) = 2^x(4)

2^2x – 2⁰ = 2^x(4)

2^2x – 1 = 2^x(4)

Then use the power of a power law to the term 2^2x:

(2^x)² – 1 = 2^x(4)

And then write in standard form:

(2^x)² – 4(2^x) – 1 = 0

Then you apply the quadratic formula, leaving you with 2 + sqrt(5) and 2 - sqrt(5).

Alternatively, you could have set k = 2^x and then solved for k which would have yielded

2^x = 2 + sqrt(5) and

2^x = 2 - sqrt(5)

Which would appear to have 2 answers but after taking the common logarithm of both sides, applying the power law of logarithms, and dividing both sides by log2, you would be left with an expression that when inputted into the calculator yields approximately 2.08. As for 2^x = 2 - sqrt(5), the answer would be less than 0, which is not possible since powers always have to be a positive number so this answer can be omitted.

That's about it. Thanks for reading.