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| a. | \(\cos (A-B)-\cos (A+B)\) | \(=\cos A \cos B+\sin A \sin B-(\cos A \cos B-\sin A \sin B)\) |
| \(=2 \sin A \cos B\). |
b. \(\text{Proof (See worked solutions)}\)
| a. | \(\cos (A-B)-\cos (A+B)\) | \(=\cos A \cos B+\sin A \sin B-(\cos A \cos B-\sin A \sin B)\) |
| \(=2 \sin A \cos B\). |
b. \(\text{RTP:} \ \ \sin \theta+\sin (3 \theta)+\ldots+\sin ((2 n-1) \theta)=\dfrac{\sin ^2(n \theta)}{\sin \theta}\)
\(\text{If} \ \ n=1:\)
\(\text{LHS}=\sin \theta\)
\(\text {RHS}=\dfrac{\sin ^2 \theta}{\sin \theta}=\sin \theta=\text{LHS}\)
\(\therefore \text {True for} \ \ n=1\)
\(\text{Assume true for} \ \ n=k:\)
\(\sin \theta+\sin (3 \theta)+\ldots+\sin \left((2 k-1) \theta\right)=\dfrac{\sin ^2(k \theta)}{\sin \theta}\ \ …\ \text{(*)}\)
\(\text{Prove true for} \ \ n=k+1:\)
\(\sin \theta+\sin (3 \theta)+\cdots+\sin ((2 k-1) \theta)+\sin ((2 k+1) \theta)=\dfrac{\sin ^2((k+1) \theta)}{\sin \theta}\)
| \(\text {LHS}\) | \(=\dfrac{\sin ^2(k \theta)}{\sin \theta}+\sin ((2 k+1) \theta) \quad \text{(see (*) above)}\) |
| \(=\dfrac{\sin ^2(k \theta)+\sin \theta \times \sin ((2 k+1) \theta)}{\sin \theta}\ \ …\ (1)\) |
\(\text{Using part a:}\)
| \(\sin \theta \cdot \sin ((2 k+1) \theta)\) | \(=\dfrac{1}{2}[\cos (2 k \theta)-\cos ((2 k+2) \theta)]\) |
| \(=\dfrac{1}{2}[\cos (2 k \theta)-\cos (2(k+1) \theta)]\) | |
| \(=\dfrac{1}{2}\left[1-2 \sin ^2(k \theta)-\left(1-2 \sin ^2((k+1) \theta)\right)\right]\) | |
| \(=\dfrac{1}{2}\left(2 \sin ^2((k+1) \theta)-2 \sin ^2(k \theta)\right)\) | |
| \(=\sin ^2((k+1) \theta)-\sin ^2(k \theta)\) |
\(\text{Substitute into (1):}\)
| \(\text {LHS}\) | \(=\dfrac{\sin ^2(k \theta)+\sin ^2((k+1) \theta)-\sin ^2(k \theta)}{\sin \theta}\) |
| \(=\dfrac{\sin ^2((k+1) \theta)}{\sin \theta}=\text{RHS}\) |
\(\Rightarrow\ \text{True for}\ \ n=k+1\)
\(\therefore\ \text{Since true for}\ \ n=1, \text{by PMI, true for integers}\ \ n \geq 1.\)