Simplify the trigonometric expression

Question

 cotxsinx-sin((π/2)-x) +cosx

Answer 1

First, since cot(x) ≡ cos(x)/sin(x), your first term is a convoluted way of saying cos(x), because sin(x) is both in numerator and denominator.

 

You can expand sin[(n/2) - x] to sin(n/2)cos(x) - cos(n/2)sin(x), but this doesn't help much...it's not what I'd call a "simplification" but more of an "expansion."

 

So just add your two cos(x) terms together and you have

 

cot(x)sin(x) - sin[(n/2) - x] + cos(x) = 2cos(x) - sin[(n/2) - x]

Answer 2

cotxsinx - sin(π/2 - x) + cosx

 

= (cosx/sinx) sinx -[sin(π/2)cosx - cos(π/2)sinx] + cosx

 

= cosx - [(1)(cosx) - (0)sinx] + cosx

 

= cosx - cosx + cosx = cosx