There is a group of Trig Identities that contain: #cos(A-B)=cos(A)cos(B)+sin(A)sin(B)# For your question this translates to: #cos(x-pi/2)=cos(x)cos(pi/2)+sin(x)sin(pi/2)#
For cos pi, the angle pi lies on the negative x-axis. Thus, cos pi value = -1. Since the cosine function is a periodic function, we can represent cos pi as, cos pi = cos (pi + n × 2pi), n ∈ Z. ⇒ cos pi = cos 3pi = cos 5pi , and so on. Note: Since, cosine is an even function, the value of cos (-pi) = cos (pi). So we have. cos ( x) sin ( x) If we multiply it by two we have. 2 cos ( x) sin ( x) Which we can say it's a sum. cos ( x) sin ( x) + sin ( x) cos ( x) Which is the double angle formula of the sine. cos ( x) sin ( x) + sin ( x) cos ( x) = sin ( 2 x) But since we multiplied by 2 early on to get to that, we need to divide by two to make the First, we isolate the trig expression: 16 cos ( 15 x) + 8 = 2 16 cos ( 15 x) = − 6 cos ( 15 x) = − 0.375. Use the calculator and round to the nearest thousandth: cos − 1 ( − 0.375) = 1.955. Use the identity cos ( θ) = cos ( − θ) to find that the second solution within [ − π, π] is − 1.955 . sCLG2ex.