7sin2 θ + 3 cos2θ = 4 show that tan θ = 1/√3

Expert Answers

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

`sin2theta `

`= 2sintheta*costheta`

`= (2sintheta*costheta)/1`

`= (2sintheta*costheta)/(sin^2theta+cos^2theta)`

`= (2sintheta*costheta)/(cos^2theta((sin^2theta)/(cos^2theta)+sin^2theta))`

`= (2sintheta/costheta)/((sin^2theta/cos^2theta)+1)`

`= (2tantheta)/(1+tan^2theta)`

In the same manner you can derive a term for `cos2theta` using tantheta.

`cos2theta`

`= cos^2theta-sin^2theta`

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

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

.

.

.

`= (1-tan^2theta)/(1+tan^2theta)`

`7sin2theta + 3 cos2theta = 4`

`(7*2tantheta)/(1+tan^2theta)+(3(1-tan^2theta))/(1+tan^2theta)= 4`

...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

`sin2theta `

`= 2sintheta*costheta`

`= (2sintheta*costheta)/1`

`= (2sintheta*costheta)/(sin^2theta+cos^2theta)`

`= (2sintheta*costheta)/(cos^2theta((sin^2theta)/(cos^2theta)+sin^2theta))`

`= (2sintheta/costheta)/((sin^2theta/cos^2theta)+1)`

`= (2tantheta)/(1+tan^2theta)`

 

In the same manner you can derive a term for `cos2theta` using tantheta.

`cos2theta`

`= cos^2theta-sin^2theta`

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

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

.

.

.

`= (1-tan^2theta)/(1+tan^2theta)`

 

 

`7sin2theta + 3 cos2theta = 4`

`(7*2tantheta)/(1+tan^2theta)+(3(1-tan^2theta))/(1+tan^2theta)= 4`

`14tantheta+3-3tan^2theta = 4(1+tan^2theta)`

`14tantheta+3-3tan^2theta = 4+4tan^2theta`

`7tan^2theta-14tantheta+1 = 0`

 

If `tantheta = t`

 

`7t^2-14t+1 = 0`

The solution for this type of quadratic equation is given by;

`t = (14+-sqrt(14^2-4*1*7))/(2*7)`

 

t = 1.93 OR t = 0.0742

 

Therefore;

`tantheta = 1.93`

`tantheta = 0.0742`

 

The value given in the answer is wrong.

 

Justification for error is given below.


If `tantheta = 1/sqrt3 then theta = 30`

 

`7sin2theta + 3 cos2theta = 4`

 

LHS

`7sin60 + 3cos60`

`= 7*sqrt3/2+3*1/2`

`= (7sqrt3+1)/2`

 

RHS = 4

 

`LHS != RHS `

 

 

 

 

Approved by eNotes Editorial Team