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



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violy's profile pic

Posted on (Answer #1)

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).


oldnick's profile pic

Posted on (Answer #3)






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

mjripalda's profile pic

Posted on (Answer #2)

`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)` .



Then, simplify the expression inside the parenthesis.



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




Rationalize the denominator to simplify further.


`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.

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