14 Use \( \cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta \) to prove that. 1) \( \cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta \) iii) \( \sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta \) iii) \( \sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta \) I. 5 Use the above to prove that i) \( \sin 2 \beta=2 \sin \beta \cos \beta \) ii) \( \cos 2 \beta=1-2 \sin ^{2} \beta \) 1.6 Evaluat \( \sin 75^{\circ} \cos 75^{\circ} \) Without using a calculator.

**1. Prove: \( \cos(\alpha+\beta)=\cos\alpha\,\cos\beta-\sin\alpha\,\sin\beta \)** Starting with \[ \cos(\alpha-\beta)=\cos\alpha\,\cos\beta+\sin\alpha\,\sin\beta, \] replace \(\beta\) with \(-\beta\) to obtain an expression for \(\cos(\alpha+ \beta)\): \[ \cos(\alpha-(-\beta))=\cos\alpha\,\cos(-\beta)+\sin\alpha\,\sin(-\beta). \] Since \[ \cos(-\beta)=\cos\beta\quad \text{and}\quad \sin(-\beta)=-\sin\beta, \] this becomes \[ \cos(\alpha+\beta)=\cos\alpha\,\cos\beta-\sin\alpha\,\sin\beta. \] --- **2. Prove: \( \sin(\alpha+\beta)=\sin\alpha\,\cos\beta+\cos\alpha\,\sin\beta \)** Express \(\sin(\alpha+\beta)\) in terms of cosine by using the complementary angle: \[ \sin(\alpha+\beta)=\cos\Bigl(\frac{\pi}{2}-(\alpha+\beta)\Bigr). \] Rewrite the argument: \[ \frac{\pi}{2}-(\alpha+\beta)=\Bigl(\frac{\pi}{2}-\alpha\Bigr)-\beta. \] Now apply the cosine difference formula: \[ \cos\Bigl[\Bigl(\frac{\pi}{2}-\alpha\Bigr)-\beta\Bigr] =\cos\Bigl(\frac{\pi}{2}-\alpha\Bigr)\cos\beta+\sin\Bigl(\frac{\pi}{2}-\alpha\Bigr)\sin\beta. \] Recall the co-function identities: \[ \cos\Bigl(\frac{\pi}{2}-\alpha\Bigr)=\sin\alpha,\quad \sin\Bigl(\frac{\pi}{2}-\alpha\Bigr)=\cos\alpha. \] Thus, \[ \sin(\alpha+\beta)=\sin\alpha\,\cos\beta+\cos\alpha\,\sin\beta. \] --- **3. Prove: \( \sin(\alpha-\beta)=\sin\alpha\,\cos\beta-\cos\alpha\,\sin\beta \)** A similar approach can be used. Write: \[ \sin(\alpha-\beta)=\cos\Bigl(\frac{\pi}{2}-(\alpha-\beta)\Bigr) =\cos\Bigl[\Bigl(\frac{\pi}{2}-\alpha\Bigr)+\beta\Bigr]. \] Now use the cosine addition formula: \[ \cos\Bigl[\Bigl(\frac{\pi}{2}-\alpha\Bigr)+\beta\Bigr] =\cos\Bigl(\frac{\pi}{2}-\alpha\Bigr)\cos\beta-\sin\Bigl(\frac{\pi}{2}-\alpha\Bigr)\sin\beta. \] Substitute the co-function identities: \[ \cos\Bigl(\frac{\pi}{2}-\alpha\Bigr)=\sin\alpha,\quad \sin\Bigl(\frac{\pi}{2}-\alpha\Bigr)=\cos\alpha. \] Thus, \[ \sin(\alpha-\beta)=\sin\alpha\,\cos\beta-\cos\alpha\,\sin\beta. \] --- **4. Prove the Double-Angle Formulas** *(i) Prove: \( \sin2\beta=2\sin\beta\,\cos\beta \)* Use the sine addition formula with \(\alpha=\beta\): \[ \sin2\beta=\sin(\beta+\beta) =\sin\beta\,\cos\beta+\cos\beta\,\sin\beta =2\sin\beta\,\cos\beta. \] *(ii) Prove: \( \cos2\beta=1-2\sin^2\beta \)* Use the cosine addition formula with \(\alpha=\beta\): \[ \cos2\beta=\cos(\beta+\beta) =\cos\beta\,\cos\beta-\sin\beta\,\sin\beta =\cos^2\beta-\sin^2\beta. \] Recall the Pythagorean identity \(\cos^2\beta=1-\sin^2\beta\). Substitute: \[ \cos2\beta=(1-\sin^2\beta)-\sin^2\beta=1-2\sin^2\beta. \] --- **5. Evaluate: \( \sin75^\circ \cos75^\circ \) Without a Calculator** Use the double-angle formula for sine: \[ \sin75^\circ \cos75^\circ = \frac{1}{2}\sin(2\times75^\circ)=\frac{1}{2}\sin150^\circ. \] Since \[ \sin150^\circ=\sin(180^\circ-30^\circ)=\sin30^\circ=\frac{1}{2}, \] it follows that \[ \sin75^\circ \cos75^\circ = \frac{1}{2}\times\frac{1}{2}=\frac{1}{4}. \]