If $cos (65^{0} - A) cos (25^{0} + B) - sin (65^{0} - A) sin (25^{0} + B) = sin (m + A - B)$ .Find $m$

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Solution

$c o s (65^{\circ} - A) c o s (25^{\circ} + B) - s i n (65^{\circ} - A) s i n (25^{\circ} + B) = c o s [(65^{\circ} - A) + (25^{\circ} + B)] = c o s [90^{\circ} - (A - B)] = s i n (A - B) n o w c o m p a r e i t w i t h g i v e n v a l u e s i n (m + A - B) w e g e t m = 0$

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f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ω^{3} & ω^{4} & ω^{5} sin (m - 1) x & sin m x & sin (m + 1) x cos (m - 1) x & cos m x & cos (m + 1) x \end{matrix} ∣ ∣ ∣ ∣ ∣

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