Question

periodic function flexibility quick check

rewrite the sine function ( f(x)=2sinleft(x+\frac{pi}{2}
ight)-1 ) as a cosine function. (1 point)

( \bigcirc f(x)=2cos(x+pi)-1 )

( \bigcirc f(x)=2cosleft(x+\frac{pi}{4}
ight)-1 )

( \bigcirc f(x)=2cos(x)-1 )

( \bigcirc f(x)=2cosleft(x-\frac{pi}{2}
ight)-1 )

Explanation:

Step1: Use the trigonometric identity

Recall the identity $\sin(A + B)=\sin A\cos B+\cos A\sin B$. For $A = x$ and $B=\frac{\pi}{2}$, we have $\sin(x+\frac{\pi}{2})=\sin x\cos\frac{\pi}{2}+\cos x\sin\frac{\pi}{2}$. Since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2}=1$, then $\sin(x+\frac{\pi}{2})=\cos x$.

Step2: Substitute the identity into the function

Given $f(x)=2\sin(x + \frac{\pi}{2})-1$, substituting $\sin(x+\frac{\pi}{2})=\cos x$ gives $f(x)=2\cos x-1$.

Answer:

$f(x)=2\cos(x)-1$ (the third option)