## University Calculus: Early Transcendentals (3rd Edition)

- Apply Addition Formula for cosine to the left side. - Simplify. Then, we have the identity: $$\cos\Big(x-\frac{\pi}{2}\Big)=\sin x$$
*Addition Formulas for cosine: $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ $$\cos\Big(x-\frac{\pi}{2}\Big)=\sin x$$ *Consider the left side: $$\cos\Big(x-\frac{\pi}{2}\Big)=\cos\Big[x+\Big(-\frac{\pi}{2}\Big)\Big]$$ Apply Addition Formulas here: $$\cos\Big(x-\frac{\pi}{2}\Big)=\cos x\cos\Big(-\frac{\pi}{2}\Big)-\sin x\sin\Big(-\frac{\pi}{2}\Big)$$ $$\cos\Big(x-\frac{\pi}{2}\Big)=\cos x\times0-\sin x\times(-1)$$ $$\cos\Big(x-\frac{\pi}{2}\Big)=\sin x$$ The identity has been proved.