Sum and Difference Formulas
\(\cos \left( {a – b} \right) = \cos a\cos b + \sin a\sin b\)
\(\cos \left( {a + b} \right) = \cos a\cos b – \sin a\sin b\)
\(\sin \left( {a – b} \right) = \sin a\cos b – \cos a\sin b\)
\(\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\)
\(\tan \left( {a – b} \right) = \dfrac{{\tan a – \tan b}}{{1 + \tan a\tan b}}\)
\(\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 – \tan a\tan b}}\)
Application Exercises of Sum and Difference Formulas
Example 1: Given \(\sin x = \dfrac{1}{2}, 0 < x < \dfrac{\pi}{2}\). Calculate the value of \(\cos \left( {x + \dfrac{\pi}{4}} \right)\).
Solution:
Since \(0 < x < \dfrac{\pi}{2}\), the angle lies in the first quadrant \(\Rightarrow \cos x > 0 \Rightarrow \cos x = \dfrac{{\sqrt 3}}{2}\).
We have \(\cos \left( {x + \dfrac{\pi}{4}} \right) = \cos x \cdot \cos \dfrac{\pi}{4} – \sin x \cdot \sin \dfrac{\pi}{4} = \dfrac{{\sqrt 2}}{2} \cos x – \dfrac{{\sqrt 2}}{2} \sin x = \dfrac{{\sqrt 6 – \sqrt 2}}{4}\).
Example 2: Simplify the expression \(A = \sin \left( {x + 14^\circ} \right) \sin \left( {x + 74^\circ} \right) + \sin \left( {x – 76^\circ} \right) \sin \left( {x – 16^\circ} \right)\).
Solution:
We have \(A = \sin \left( {14^\circ + x} \right) \cos \left( {16^\circ – x} \right) + \sin \left( {76^\circ – x} \right) \sin \left( {16^\circ – x} \right)\)
\( = \sin \left( {14^\circ + 16^\circ + x – x} \right) = \sin 30^\circ = \dfrac{1}{2}\).
Example 3: Without using a calculator, calculate the following trigonometric values: \(\cos {795^\circ}\), \(\tan \dfrac{7\pi}{12}\).
Solution:
* Calculate \(\cos {795^\circ}\)
\(\begin{array}{l}\cos {795^\circ} = \cos \left( {{75^\circ} + {2 \cdot 360^\circ}} \right)\\ = \cos {75^\circ} = \cos \left( {{30^\circ} + {45^\circ}} \right)\\ = \cos {30^\circ} \cos {45^\circ} – \sin {30^\circ} \sin {45^\circ}\\ = \dfrac{{\sqrt 3}}{2} \cdot \dfrac{{\sqrt 2}}{2} – \dfrac{1}{2} \cdot \dfrac{{\sqrt 2}}{2} = \dfrac{{\sqrt 6 – \sqrt 2}}{4}\end{array}\)
* Calculate \(\tan \dfrac{7\pi}{12}\)
\(\begin{array}{l}\tan \dfrac{7\pi}{12} = \tan \left( {\dfrac{\pi}{3} + \dfrac{\pi}{4}} \right)\\ = \dfrac{\tan \dfrac{\pi}{3} + \tan \dfrac{\pi}{4}}{1 – \tan \dfrac{\pi}{3} \tan \dfrac{\pi}{4}} = \dfrac{\sqrt 3 + 1}{1 – \sqrt 3} = -2 – \sqrt 3 \end{array}\)
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