Trig Identities:

a) sin^4 x - cos^4 x / sin^2 x - cos^2 x = 1 

b) 1-2 cos^2 x = tan^2 x-1/tan^2 x+1 

c) secx + tanx = cosx/1 - sinx

 

Most of these identities hold true but some of them are false. Prove LS = RS or show why the rule does not hold

Expert Answers

An illustration of the letter 'A' in a speech bubbles

a) `(sin^4(x) - cos^4(x))/(sin^2(x)-cos^2(x)) = 1`

LHS:

`((sin^2(x) - cos^2(x))(sin^2(x)+ cos^2(x)))/(sin^2(x)-cos^2(x))`

We know, from basic trignometry,

sin^2(x)+ cos^2(x) = 1

Therefore,

`LHS=((sin^2(x) - cos^2(x))(1))/(sin^2(x)-cos^2(x))`

LHS = 1.

Therefore LHS = RHS and

`(sin^4(x) - cos^4(x))/(sin^2(x)-cos^2(x)) = 1`

b) `(1-2cos^2(x)) = (tan^2(x)-1)/(tan^2(x)+1)`

For this I will start with RHS.

`RHS = (tan^2(x)-1)/(tan^2(x)+1)`

`= ((sin^2(x))/(cos^2(x))-1)/((sin^2(x))/(cos^2(x))+1)`

`= (sin^2(x)-cos^2(x))/(sin^2(x)+cos^2(x))`

`= (sin^2(x)-cos^2(x))/1`

`= (1-cos^2(x))-cos^2(x)`

`= 1-2cos^2(x)`

Therefore LHS = RHS and

`(1-2cos^2(x)) = (tan^2(x)-1)/(tan^2(x)+1)`

c) `sec(x)+tan(x) = cos(x)/(1-sin(x))`

`LHS = sec(x)+tan(x)`

`= 1/cos(x) + sin(x)/cos(x)`

`=(1+sin(x))/cos(x)`

Multiplying both numerator and denominator by `(1-sin(x))`

`= (1+sin(x))(1-sin(x))/(cos(x)(1-sin(x)))`

`= (1-sin^2(x))/(cos(x)(1-sin(x)))`

`= (cos^2(x))/(cos(x)(1-sin(x)))`

`= cos(x)/(1-sin(x))`

LHS = RHS and

`sec(x)+tan(x) = cos(x)/(1-sin(x))`

All the identities are true.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial