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How solve trigonometric equation sinx+cosx=1?

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yapayapa | Honors

Posted September 4, 2013 at 5:05 PM via web

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How solve trigonometric equation sinx+cosx=1?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 4, 2013 at 5:17 PM (Answer #1)

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You need to use the following trigonometric identities, such that:

`sin x = (2tan(x/2))/(1 + tan^2(x/2))`

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

Replacing `(2tan(x/2))/(1 + tan^2(x/2))` for `sin x` and `(1 - tan^2(x/2))/((1 + tan^2(x/2)))` for `cos x` in equation, yields:

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

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

You should come up with the following substitution, such that:

`tan(x/2) = t`

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

Factoring out `2t` yields:

`2t(t - 1) = 0`

Using the zero product rule yields:

`2t = 0 => t = 0`

`t - 1 = 0 => t = 1`

You need to solve for x the equation `tan(x/2) = t` , such that:

`tan(x/2) = 0 => x/2 = tan^(-1)0 + n*pi`

`x/2 = 0 + n*pi => x = 2n*pi`

`tan(x/2) = 1 => x/2 = tan^(-1)(1) + n*pi`

`x/2 = pi/4 + n*pi => x = (2pi)/4 + 2n*pi => x = pi/2 + 2n*pi`

Hence, evaluating the general solutions to the given trigonometric equation, yields `x = 2n*pi` and `x = 2n*pi + pi/2.`

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