Trigonometry help required (1 Viewer)

[movie_detective]

Hey all,

For this question: sec(2A) + tan(2A)= cos(A) + sin(A) /cos(A) -sin(A), I'm just wondering if it is possible to obtain the answer without cross multiplying at one stage or another ? e.g only working on the left side .

Thx for helping

 

[combflake]

 

[movie_detective]

Thx, nice working out. One aspect I noticed in the solution was that in order to answer my question by working on one side is to work backwards e.g 1= sin^2x + cos^2x . I rarely get questions which expect you to do that ...

 

[combflake]

Instead of multiplying the top and bottom of the 2nd term by cos^2 A, you can instead DIVIDE the top and bottom of the FIRST term by cos^2 A.
After replacing the resulting sec^2 A by (1+tan^2 A), you get another perfect square in the numerator which cancels as before.
This avoids magically turning 1 into sin^2 A + cos^2 A.

 

[movie_detective]

Thx, just another 2 quick questions which link:Is it possible to rationalize the right hand side ? And if a question has a denominator like Sinx-cosx or Sinx-1 , are they able to be rationalized ?

 

[BrightFella]

There are two simple identities usually taught at higher 3u/4u from which you can simplify further:

1. cos(u)+cos(v)=2cos[(u+v)/2]cos[(u-v)/2]
2. cos(u)-cos(v)=-2sin[(u+v)/2]sin[(u-v)/2]

Let v=pi/2-u. Then we get:

1. cos(u)+sin(u)=2cos[pi/4]cos[(-pi/2+2u)/2]
2. cos(u)-sin(u)=-2sin[pi/4]sin[-(pi/2+2u)/2]

Hence dividing top by bottom you get beautiful result in terms of A:

(cosA+sinA)/(cosA-sinA)=-cot[(-pi/2+2A)/2]=cot(pi/4-A)