29+Proving+Trigonometric+Identities

= Proving Trigonometric Identites = By: Emma Wieduwilt Period 1

I chose the topic of proving trigonometric identities because I enjoyed using formulas we learned to substitute other formulas in order to prove an equality. This was a bit difficult for me and sometimes I had to step away from the problem and try to see it from a different point of view because there are so many different ways to prove an identity. Trigonometric Identities: secx=1/cosx or cosx=1/secx cscx=1/sinx or sinx=1/cscx cotx=1/tanx or tanx=1/cotx sinx^2x+cos^2x=1 tan^2x+1=sec^2x 1+cot^2x=csc^2x tanx=sinx/cosx cotx=cosx/sinx
 * __Background:__**
 * __Explanation:__**

A typical problem given asks for you to prove that an equation is true by substituting the identities for a given expression. -transform the more complicated side of the equation into the form of the simpler side -substitute one or more basic trigonometric identities to simplify the expression -factor or multiply to simplify the expression -least common denominator -multiply numerator and denominator by the same trigonometric expression -difference of squares
 * Some tips:**


 * __Examples:__** **bold**= identity used cos/sin/tan^2x=cos/sin/tan^2(x)


 * Easy: **

1)Simplifying an expression: cosxtanxcscx cosx*sinx/cosx *1/sinx >>>**cosines and sines cancel so you are left with 1**
 * cosx=cosx tanx=sinx/cosx cscx=1/sinx**

2) 1/sec^2x + 1/csc^2x = 1 (solving left side) cos^2x + sin^2x =**1** is one of the identities so it works!
 * 1/sec^2x =cos^2x 1/csc^2x=sin^2x**


 * Medium: **

3) secx/cosx - tanx/cotx = 1 (solving left side) sec^2x - tanx/cotx sec^2x - (sinx/cosx)/(cosx/sinx) sec^2x - sin^2x/cos^2x sec^2x - tan^2x = **1** because it's an identity
 * 1/cosx=secx**
 * tanx= sinx/cosx** and **cotx= cosx/sinx**
 * sin^2x/cos^2x= tan^2x**

4) tan^2x (cos^2x) = 1 - cos^2x (solving left side) (sin^2x/cos^2x) * cos^2x >>>**cosines cancel** sin^2x = 1 - cos^2x 1-cos^2x= 1-cos^2x
 * tan^2x = sin^2x/cos^2x**
 * sinx^2x+cos^2x=1 so 1- cos^2x = sin^2x**


 * Hard: **

5) sec^4x - sec^2x = 1/cot^4x + 1/cot^2x (solving right side) tan^4x + tan^2x tan^2x(tan^2x + 1) >>>**factor** tan^2x(sec^2x) >>**substitute** (sec^2x-1) (sec^2x) >>**distribute** sec^4x-sec^2x = sec^4x - sec^2x
 * cotx=1/tanx or tanx=1/cotx**
 * tan^2x+1=sec^2x**

6) (1+tanx) / (1 +cotx) = sinx / cosx (solving left side) (1 + (sinx/cosx)) / (1+ (cosx/sinx) >> multiply 1 in numerator by cosx/cosx and multiply numerator in denominator by sinx/sinx >>to **get common denominator** ( (cosx/cosx) + (sinx/cosx) ) / ( (sinx/sinx) + (cosx/sinx) ) >>> **add the fractions** ((cosx+sinx)/cosx) / ( (sinx+cosx)/sinx ) >>>multiply by reciprical ((cosx+sinx)/cosx) * (sinx/(sinx+cosx)) >>> **cosx+sinx and sinx+cosx cancel** (best if shown as numerator over denominator) you're left with sinx/cosx = sinx/cosx
 * tanx= sinx/cosx** and **cotx= cosx/sinx** >>**substitute**

__**Video:**__ media type="file" key="Proving Trigonometric Functions Math Video.mp4" width="546" height="546" __**[|Proving Trigonometric Functions Math Video.mp4]**__ []

__**Class Problem:**__ Prove that: 1 - cot^4x = 2csc^2x - csc^4x