Simplify sin(2x + pi/2) + 5.

Step-by-step solution

1. 1

   Identify the identity to use. A quarter-turn phase shift converts sine into cosine:

   $\sin\!\left(\theta+\frac{\pi}{2}\right)=\cos\theta$

   This follows from the addition formula: $\sin\theta\cos\tfrac{\pi}{2}+\cos\theta\sin\tfrac{\pi}{2}=\sin\theta\cdot0+\cos\theta\cdot1=\cos\theta$.

2. 2

   Match the pattern correctly. Here the whole inner expression is $2x+\tfrac{\pi}{2}$, so $\theta=2x$ — the entire $2x$, not just $x$. Substituting:

   $\sin\!\left(2x+\frac{\pi}{2}\right)=\cos(2x)$

   A frequent error is writing $\cos\!\left(x+\tfrac{\pi}{2}\right)$ or $2\cos x$ instead.

3. 3

   Leave the constant alone. The $+5$ sits outside the trig function, so no identity touches it:

   $\sin\!\left(2x+\frac{\pi}{2}\right)+5=\cos(2x)+5$

4. 4

   Read off the properties of the simplified form. As $\cos(2x)+5$, the function has amplitude $1$, period $\tfrac{2\pi}{2}=\pi$, midline $y=5$, and range $[4,6]$. All of this is far easier to see after the simplification than before it — which is the point of doing it.

5. 5

   Note that the simplification also removed the phase shift. The original had a horizontal shift of $-\tfrac{\pi}{4}$ in $x$ (since $2x+\tfrac{\pi}{2}=2\left(x+\tfrac{\pi}{4}\right)$); rewriting as a cosine absorbs it entirely, leaving a phase shift of zero.

6. 6

   Verify numerically. Comparing $\sin\!\left(2x+\tfrac{\pi}{2}\right)+5$ with $\cos(2x)+5$ at $40$ random values of $x$ drawn from $[-9,9]$ gives agreement to machine precision at every point ✓.