Q.$\cos(3-2x)$

Let $y=\cos(3-2x)$. Since cosine is an even function, $\cos(3-2x)=\cos[-(2x-3)]=\cos(2x-3)$.

So $y=\cos(2x-3)$, which is now in the standard form $\cos(ax+b)$ with $a=2$, $b=-3$.

Step 1 — differentiate directly to confirm the pattern: $y_1=-2\sin(2x-3)=2\sin\!\left(-(2x-3)\right)$... instead use $-\sin\theta=\cos(\theta+\pi/2)$ directly: $y_1=2\cos\!\left(2x-3+\dfrac{\pi}{2}\right)$.

Step 2: $y_2=-4\cos(2x-3)=4\cos(2x-3+\pi)$.

Step 3: $y_3=8\sin(2x-3)=8\cos\!\left(2x-3+\dfrac{3\pi}{2}\right)$. …

Use $\cos(-\theta)=\cos\theta$ to rewrite $\cos(3-2x)$ as $\cos(2x-3)$ FIRST, then apply the standard $\cos(ax+b)$ n …

Applying the pattern directly to $\cos(3-2x)$ with '$a=-2$' without first using the evenness of cosine to flip the sign — this i …