What Are Some Common Trigonometric Simplifications and Derivative Rules?

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SUMMARY

This discussion focuses on common trigonometric simplifications and derivative rules, specifically addressing problems involving the simplification of expressions like 2 sin(v) cos(v) - sin(v) and the application of the chain rule in differentiation. Participants clarify that cos^2(v) - sin^2(v) can be rewritten as 2cos^2(v) - 1 using the identity for cosine of double angles. Additionally, the discussion highlights the importance of understanding the behavior of the tangent function and its intervals, particularly when solving inequalities like tan(x) < 1.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cos^2(v) + sin^2(v) = 1
  • Familiarity with the chain rule in calculus, expressed as (u^n)' = n*u' * u^(n-1)
  • Knowledge of how to manipulate and simplify algebraic expressions involving trigonometric functions
  • Basic understanding of the properties of the tangent function and its periodicity
NEXT STEPS
  • Research the derivation and application of the double angle formulas for sine and cosine
  • Learn about the implications of the chain rule in more complex differentiation problems
  • Study the behavior of trigonometric functions over different intervals, focusing on their increasing and decreasing properties
  • Explore advanced techniques for solving trigonometric inequalities, including graphical methods
USEFUL FOR

Students preparing for exams in trigonometry and calculus, educators teaching these subjects, and anyone looking to deepen their understanding of trigonometric identities and differentiation techniques.

pace
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different (easy) math questions before my exam :)

1.

1. Homework Statement :

Write as simple as possible: 2 sinv cosv-sinv
------------------
cos^2v-sin^2v-cosv+1

(cos^2v same as (cosv)^2)


2. Homework Equations :

sinv
tanx= -----
cosv

cos^2v+sin^2v = 1







3.

Sollution my previous helper says:
cos^2v-sin^2v is supposed to be 2cos^2v - 1.
Because: cos2x=cos^v-sin^2v (?)
=2cos^2-1 ((and solves it easy after that))


How's that?


My attempt at a solution:

2 sinv cosv-sinv
------------------
cos^2v-sin^2v-cosv+1

sinv(2cosv-1)
----------------
cos^2v-(1-cos^2v) -cosv+1


sinv(2cosv-1)
----------------
2cos^2v-cos v

... and then I don't know what to do. I can't just delete one cosv in each part because it's one 2cos^2v there, and it's different..

[/b]


2.

1. Homework Statement :

This is inside a test ((64+t^2)^2)` it's supposed to be: 2(64+t^2) x 2t.



2. Homework Equations :

My helped reffered to that (u^n)` = n*u` * u^n-1. Which gives the above result. I don't see that law in my 2mx book, is it 3mx? But this is a 2mx test. My book though states that (x^n)`= n*u^n-1 (eg: (-6x^2)`= -12x) , but that doesn't go for biggies like (64+t^2), am I right?


3. The Attempt at a Solution :

My main question is just where is that law from?



3.

1. Homework Statement :

When do you use {and ( together, like: {(u)} ?




4.

1. Homework Statement & Relevant equations & The attempt at a solution:

My book states that 3[squareroot](-5)^3 = -5, and that 4[squareroot](-5)^4 = 5, and so on in odd and even numbers. But laters say that 4[squareroot]a^4, when a is a negative number(as in -5 above) gets to be -a. As if 4[squareroot](-5)^4 = 5, but even numbers should give positive results. Why does it change?




5.

1. Homework Statement

How do you solve tanx<1 nicely on paper?

The problem statement is tanx<1


2. Homework Equations :

sinv
tanx= -----
cosv

tan(v+k 180') = tanv



3. The Attempt at a Solution

tanx<1
x<45'(degrees) and x<225'(degrees)
L=<45',0'> and <90',225'> and <270,360'>

But my counting seems so short. Is there any rules that say that I'm writing too short, then if so what am I missing?



All help appreciated! :) Hope I placed this at the right place O_o
 
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First of all, if you start clicking on "fancy equations" that other people have entered in other threads, a little text box will come up to show how to enter your equations in LaTeX. There's also a tutorial here: https://www.physicsforums.com/showthread.php?t=8997 It's pretty easy to do; I'm surprised you don't already do it (given 143 posts) It makes it much nicer to read :)

On #1, you can factor the denominator the same way you factored the numerator originally: there's a greatest common factor.

On #4, try plugging in a few negative numbers, raise them to the 4th power, then taking the 4th root.
 
pace said:
This is inside a test ((64+t^2)^2)` it's supposed to be: 2(64+t^2) x 2t. [/b]


2. Homework Equations :

My helped reffered to that (u^n)` = n*u` * u^n-1. Which gives the above result. I don't see that law in my 2mx book, is it 3mx? But this is a 2mx test. My book though states that (x^n)`= n*u^n-1 (eg: (-6x^2)`= -12x) , but that doesn't go for biggies like (64+t^2), am I right?


3. The Attempt at a Solution :

My main question is just where is that law from?

The law you are referring to: [tex]\frac{d}{dx} u^n = (nu^{n-1}) \cdot \frac{d}{dx} u[/tex], comes from the more general result differentiation, the chain rule. This rule states that : [tex]\frac{dy}{du} \cdot \frac{du}{dx} = \frac{dy}{du}[/tex].

3.

1. Homework Statement :

When do you use {and ( together, like: {(u)} ?

Do you mean in terms of mathematical notation or LaTeX? LaTeX you sometimes need it, but you don't use it anyway. In mathematics, the normal curved brackets can either denote the argument of a function, ie f(x), g(t), or to clarify products, eg [itex](23-x)(23+x)[/itex].

4.

1. Homework Statement & Relevant equations & The attempt at a solution:

My book states that 3[squareroot](-5)^3 = -5, and that 4[squareroot](-5)^4 = 5, and so on in odd and even numbers. But laters say that 4[squareroot]a^4, when a is a negative number(as in -5 above) gets to be -a. As if 4[squareroot](-5)^4 = 5, but even numbers should give positive results. Why does it change?
It seems when you say "3[squareroot]" you mean the 3rd root, not 3 times the root, please clarify that next time. And they made a mistake which is what is causing your confusion, the 4th root of a to the power of 4 is positive a. However they may be trying to find a solution to something, and just like in the quadratic equation, both the positive and negative roots must be taken into account, so -a is one of the other roots to be considered.

5.

1. Homework Statement

How do you solve tanx<1 nicely on paper?

The problem statement is tanx<1


2. Homework Equations :

sinv
tanx= -----
cosv

tan(v+k 180') = tanv



3. The Attempt at a Solution

tanx<1
x<45'(degrees) and x<225'(degrees)
L=<45',0'> and <90',225'> and <270,360'>

But my counting seems so short. Is there any rules that say that I'm writing too short, then if so what am I missing?



All help appreciated! :) Hope I placed this at the right place O_o

Ok well it seems that you want to find all values of x between 0 and 360 degrees where tan x is less than 1. To aid you, note that tan has a period of 180 degrees, so you only need to find values where tan x < 1 between 0 and 180 degrees, then add on 180 degrees to those solutions after. Between 0 and 180 degrees, we know that tan x = 1 when x is 45 degrees. We also know that the tan function is in increasing function. So all values between 0 and 45 degrees are less than 1. After 45, and up to almost 90, it is increasing. Going to 90 from left to right, tan goes to positive infinity, much bigger than 1. But from just after 90, it starts at negative infinity again, so these values are allowed. Then we know tan x is 1 at 225 again, and isn't 1 again. Since we can't include the value 90 or 225, but they are the endpoints of the values of we want, we write the interval as (90 degrees, 225 degrees) as opposed to the closed interval [90 degrees, 225 degrees].
 

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