# Is `cos(pi/2-x)= cos(pi/2)-cos(x)` ?

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It is possible to write `cos(pi/2 - x) = sin (x)` as the two trigonometric functions are shifted along the x-axis by `pi/2`

`cos (pi/2) = 0`

`cos(pi/2) - cos x = -cos x = cos x`

The right hand side is equal to cos x while the left hand side is equal to sin x.

This shows that `cos(pi/2-x) != cos(pi/2)-cos(x)`

You may use the following formula for `cos(alpha - beta)` to find if both sides are equal such that:

`cos(alpha - beta) = cos alpha*cos beta + sin alpha*sin beta`

Substituting `pi/2` for `alpha` and x for `beta ` yields:

`cos(pi/2 - x) = cos (pi/2)cos x + sin (pi/2)*sin x`

Notice the differences between the result above and the right side provided by the problem.

`cos (pi/2)cos x + sin (pi/2)*sin x != cos (pi/2) - cos x`

You may also try the following approach by transforming the right side into a product using the formula:

`cos alpha - cos beta = 2 sin (alpha+beta)/2* sin(beta - alpha)/2`

Hence `cos (pi/2) - cos x = 2 sin (pi/2+x)/2* sin(x - pi/2)/2`

Notice again the differences between the left side provided by the problem and the left side resulted such that:

`cos(pi/2 - x) != 2 sin (pi/2+x)/2* sin(x - pi/2)/2`

**As a conslusion, there is no identity created between both sides, hence `cos(pi/2 - x) != cos (pi/2) - cos x` .**