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Question

If $$\sin A=\frac{4}{5},\frac{\pi}{2}<A<\pi$$ and $$\cos B=\frac{5}{13},\frac{3\pi}{2}<B<2\pi$$, find 
(i) $$\sin (A+B)$$, (ii)$$\cos (A-B)$$, (iii)$$\tan (A-B)$$


Solution

$$\sin A=\cfrac{4}{5},\cos B=\cfrac{5}{13}$$
$$\sin A=\cfrac{4}{5},\ A\ \epsilon\ (\pi/2,\pi)$$
$$\implies \cos A=\cfrac{3}{\sqrt{25-16}}=\cfrac{-3}{5}$$
$$\cos B=\cfrac{5}{13}$$$$,B\ \epsilon\ (3\pi/2,2\pi)$$
So, $$\sin B=\cfrac{-12}{13}$$


$$(1)$$  $$\sin(A+B)=\sin A\cos B+\cos A\sin B\\=(\cfrac{4}{5})(\cfrac{5}{13})+(\cfrac{-3}{5})(\cfrac{-12}{13})=\cfrac{56}{65}$$

$$(2)$$  $$\cos(A-B)=\cos A\cos B+\sin A\sin B\\=(\cfrac{-3}{5})(\cfrac{5}{13})+(\cfrac{4}{5})(\cfrac{-12}{13})=\cfrac{-63}{65}$$

$$(3)$$  $$\sin(A-B)=\sin A\cos B-\cos A\sin B\\=(\cfrac{4}{5})(\cfrac{5}{13})-(\cfrac{-3}{5})(\cfrac{-12}{13})=\cfrac{-16}{65}$$


$$(4)$$  $$\tan(A-B)=\cfrac{\sin A-B}{\cos (A-B)}\\=\cfrac{(-16/65)}{(-63/65)}=\cfrac{16}{63}$$


Mathematics

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