Student Video #1-Unit S Concept 3 Assessment #3

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What is This Concept About?
This concept goes over Unit S Concept 3, which deals with using power reducing formulas to simplify trig equations. This concept is used when the trig function is at  high power, and need to be powered down in ordered to be simplified.

What is important to remember about this concept?
It is important to understand the process if substituting the formulas withing the equation in order to power down the trig functions. Also, when taking the inverse of the denominators of already complex fractions, you must be careful to multiply it correctly to the denimator of the numerator.

Student Video #2- Unit S Concept 7 Assessment #4

 
What is this concept about?
This Concept goes over solving equations with half angle formulas. there will be a trig function within the equation that is a half angle, and in order to simplify the equation, you must substitute the half angle with the corresponding formula. (Unit S Concept 7)

What is Important to remember about this concept?
What is really important about this concept is to substitute the right formula in the equation. Also, in this problem in articular, it is important to realize that in order to solve for "x", you will need to do alot of substituting, and distribution. If you do not distribute correctly, you will not be able to use the Zero Product Property correctly. Remember to keep the equation set to zero when solving for "x".

Unit S Assessment #2

 Solving Using Sum/Difference Formulas------------------------>

Solving Using Half Angle Formula             
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These two ways of finding the values of the trig functions give the same answers they are both simplifying through radical other formulas, just in a different process. The multiples within the trig functions are similar, and when you plug each of them in the calculator, they will give the same answers for each of them. Also, since we are using the half angle within the sum/difference formula, it is similar to using the "u" , but taking the square root of the values. Therefore, since each values come out the same, in know that there are two ways of finding the values of these trig function and angles.

Student Problem #10 Unit R Concept 3

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Student Problem #9 Unit R Concept 2

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Student Problem #8 Unit R Concept 1

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Concept 4 Problems Solving Trig Functions

Level 1

In this level, we simply take the trig function we have and find out how we get the given value through the unit circle. since we know that tan is x/y, we know that we can get the ratio rad3/3 with 1/2/rad3/3. the given points are found in the 60 degree, and also in 240 degree, since it is a positive function.








Level 2

The second level, we deal with simple problem solving. we must get the sin (x) by itself. In this case, we are getting no solution since sin must be -1<x<1.












Level 3
In this level, we deal with more complicated solving. we must get a common trig function, and since we know from the ratio identity, csc=1/sin, and cot=cos/sin. there is a common denominator, so we combine the equation. next, we must multiply sin to both sides. since we are now stuck, we must square both sides. we know that sin^2(x) is 1-cos^2. finally, we take what is given, and combine like terms, having a zero on the right side. we take out the common factor which is 2cos. then, we solve for both terms.
Don't forget that there may be extraneous solutions!


Level 4

In this level of problem, we are looking for half/double angles. We need to replace the function with "m", then solve for m. when we do, we take what is solve, 360 and 180, and equal it to x/4. put it as "x" and get the final angles.