Could someone help me with this question

Write in polar form : sinx - i cosx

I've come up to this point, and it's wrong, can't someone please help?

=sin(pie/2-x)-icos(pie/2-x)
=cosx-isinx
=cis(-x)

But the answer is cis(x-pie/2)

almost

sinx = cos(pi/2 - x) and cosx = sin(pi/2 - x)
that is, the sine of any angle equals the cosine of its compliment.

so sinx - i cosx
= cos(pi/2 - x) - i sin(pi/2 - x)
= cis(pi/2 - x)

this webpage has some good examples with simple diagrams to show your topic

http://www.intmath.com/Complex-numbers/4_Polar-form.php

To write the expression sin(x) - i cos(x) in polar form, we need to express it in terms of a complex exponential function. Every complex number can be written in the form a + bi, where a and b are real numbers.

In this case, sin(x) - i cos(x) can be rewritten as (-i) cos(x) + sin(x). Now, we can identify the coefficients of i and express the expression in a polar form using Euler's formula. Euler's formula states that e^(ix) = cos(x) + i sin(x).

So, by applying Euler's formula to our expression, we have:
(-i) cos(x) + sin(x) = (-i)(cos(x) + i sin(x)).

Next, let's simplify the equation:
= -i cos(x) - i^2 sin(x)
= sin(x) - i^2 cos(x)

Now, we know that i^2 is defined as -1. Substituting -1 for i^2, we get:
= sin(x) + cos(x)

This expression can further be written in polar form as follows:
= √(sin^2(x) + cos^2(x)) * (sin(x) + cos(x)) / √(sin^2(x) + cos^2(x))
= √(sin^2(x) + cos^2(x)) * [sin(x) / √(sin^2(x) + cos^2(x))] + [cos(x) / √(sin^2(x) + cos^2(x))]

Now, we can recognize that √(sin^2(x) + cos^2(x)) is equal to 1, as it represents the magnitude of this expression. Therefore, the final polar form becomes:
= 1 * [sin(x) / √(sin^2(x) + cos^2(x))] + [cos(x) / √(sin^2(x) + cos^2(x))]

Simplifying further:
= sin(x) / √(sin^2(x) + cos^2(x)) + cos(x) / √(sin^2(x) + cos^2(x))
= sin(x) / 1 + cos(x) / 1
= sin(x) + cos(x)

Hence, the polar form of sin(x) - i cos(x) is cis(x - π/2).