## Trigonometry (11th Edition) Clone

$$\cos(90^\circ-\theta)=\sin\theta$$
$$A=\cos(90^\circ-\theta)$$ Apply the cosine difference identity, which states $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ we have $$A=\cos90^\circ\cos\theta+\sin90^\theta\sin\theta$$ $$A=0\times\cos\theta+1\times\sin\theta$$ $$A=\sin\theta$$ which is actually the proof of cofunction identity for $\sin\theta$: $$\sin\theta=\cos(90^\circ-\theta)$$