# Solve this equation: sin(x + (π / 4)) – sin(x – (π / 4) = 1

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Use the identity for sum of two angles for sine, and difference of two angles for sine.

sin(a + b) = sinacosb + cosasinb and sin(a - b) = sinacosb - cosasinb

So, we will have:

`sinxcos(pi/4) +cosxsin(pi/4) - (sinxcos(pi/4) - cosxsin(pi/4)) = 1`

`sinxcos(pi/4) + cosxsin(pi/4) - sinxcos(pi/4) + cosxsin(pi/4)) = 1`

Combine like terms.

`2cosxsin(pi/4) = 1`

Substitute the value of sin(pi/4).

`2cosx(sqrt(2)/2)) = 1`

`sqrt2cosx = 1`

Isolate the cosx on left side.

`cosx = 1/sqrt(2)`

Taking the inverse cos of both sides.

**x = pi/4, 7pi/4 in [0. 2pi)**.

`sin(x+pi/4) - sin(x-pi/4)=1`

To solve, sum-to product identity of sine can be used which is `sinA-sinB=2cos((A+B)/2)sin((A-B)/2)` .

So,

`2cos(((x+pi/4)+(x-pi/4))/2)sin(((X+pi/4)-(x-pi/4))/2)=1`

Then, simplify the expression inside the parenthesis.

`2cos((2x)/2)sin((pi/2)/2)=1`

`2cosxsin(pi/4)=1`

Since `sin(pi/4)=sqrt2/2` , the equation simplifies to:

`2cosx*sqrt2/2=1`

`sqrt2cosx=1`

`cosx=1/sqrt2`

Rationalize the denominator to simplify further.

`cosx=1/sqrt2*sqrt2/sqrt2`

`cos x=sqrt2/2`

Referring to Unit Circle Chart, the values of angle x are:

`x=pi/4 , (7pi)/4`

**Since cosine repeats every 2pi, hence the general solution to the given equation are:**

**`x_1= pi/4+2pik ` and `x_2=(7pi)/4+2pik` **

**where k is any integer.**

`sin(x+pi/4)-sin(x-pi/4)=2cosxsin(pi/4)=sqrt(2)cosx`

so:

`sin(x+pi/4)-sin(x-pi/4)=1`

becames:

`sqrt(2)cosx=1`

`cosx=sqrt(2)/2` `x= pi/4+2kpi` `x=7/4pi +2kpi`