How solve equation tan (2x+pi)/4 - tan (2x-pi)/4=tanx/2+cotx/2? explain pls

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to expand the tangents to the left using the following formulas such that:

`tan (a+b) = (tan a + tan b)/(1 - tan a*tan b)`

`tan (a-b) = (tan a- tan b)/(1+ tan a*tan b)`

`tan(2x/4 + pi/4) = tan (x/2 + pi/4)`

`tan (x/2 + pi/4) = (tan(x/2) + tan (pi/4))/(1 - tan (x/2)*tan (pi/4))`

Using that `tan (pi/4) = 1`  yields:

`tan (x/2 + pi/4) = (tan (x/2)+ 1)/(1 - tan (x/2))`

`tan (x/2- pi/4) = (tan(x/2)- 1)/(1+ tan (x/2))`

Hence, the equation becomes:

`(tan(x/2) + 1)/(1 - tan (x/2)) - (tan(x/2) - 1)/(1+ tan (x/2)) = tan(x/2) + cot (x/2)`

You need to bring the terms to the left to a common denominator such that:

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

`((tan(x/2) + 1 - tan(x/2) + 1)(tan(x/2) + 1 + tan(x/2)- 1))/((1+ tan (x/2))(1 - tan (x/2))) = tan(x/2) + cot (x/2)`

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

You need to write  `cot (x/2) = 1/(tan (x/2)) ` such that:

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

You should come up with the substitution such that:

`tan (x/2) = t`

`4t = (t + 1/t)(1 - t^2)`

`4t^2 = (t^2 + 1)(1 - t^2) =gt 4t^2 = 1 - t^4`

`t^4 + 4t^2 - 1 = 0`

`t^2_(1,2) = (-4+-sqrt(16 + 4))/2`

`t^2_(1,2) = (-4+-2sqrt(5))/2`

`t^2_(1,2) = (-2+-sqrt(5))`

You need to find t such that:

`t_(1,2) = +-sqrt(-2+sqrt5)`

Since the `sqrt(-2-sqrt5) !in R` , hence, you only need to consider `t_(1,2) = +-sqrt(-2+sqrt5).`

You need to solve for x the equations:

`tan(x/2) = sqrt(-2+sqrt5) =gt x/2 = tan^(-1) (sqrt(-2+sqrt5)) + npi`

`x = 2tan^(-1) (sqrt(-2+sqrt5)) + 2npi`

`x = 2tan^(-1) (-sqrt(-2+sqrt5)) + 2npi`

Hence, evaluating the general solutions to the given equation yields that they follow the models `x = 2tan^(-1) (sqrt(-2+sqrt5)) + 2npi`  and `x = 2tan^(-1) (-sqrt(-2+sqrt5)) + 2npi.`

Sources:

We’ve answered 318,983 questions. We can answer yours, too.

Ask a question