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  • If θ lies in the first quadrant and cos θ = 8/17, then find the value of cos(30° + θ) + cos (45° – θ) + cos (120° – θ)

If θ lies in the first quadrant and cos θ = 8/17, then find the value of \cos(30 ^{\circ} + \theta ) + \cos (45 ^{\circ} - \theta ) + \cos (120 ^{\circ} - \theta )

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\\ \text{Given that } \cos \theta =\frac{8}{17}~~~ \sin \theta =\sqrt {1 - \left( \frac{8}{17} \right) ^{2}}=\sqrt {1 - \frac{64}{289}}= \sqrt {\frac{289 - 64}{289}}\\= \pm \frac{15}{17}~ \\\\ But \: \: \theta \text{ lies in I quadrant and so} \sin \theta \text{ is positive.} \sin \theta =\frac{15}{17}~ \\\\ ~Now\cos \left( 30^{0}+ \theta \right) +\cos \left( 45^{0} - \theta \right) +\cos \left( 120^{0} - \theta \right) \\\\ =\cos30^{0}\cos \theta - \sin30^{0}\sin \theta +\cos45^{0}\cos \theta +\sin45^{0}\sin \theta +\cos120^{0}\cos \theta +\sin120^{0}\sin \theta \\\\


   \\ =\frac{\sqrt {3}}{2}\cos \theta - \frac{1}{2}\sin \theta +\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta - \frac{1}{2}\cos \theta +\frac{\sqrt {3}}{2}\sin \theta \\\\ =\frac{\sqrt {3}}{2} \left( \cos \theta +\sin \theta \right) - \frac{1}{2} \left( \sin \theta +\cos \theta \right) +\frac{1}{\sqrt {2}} \left( \cos \theta +\sin \theta \right) \\\\ = \left( \frac{\sqrt {3}}{2} - \frac{1}{2}+\frac{1}{\sqrt {2}} \right) \left( \cos \theta +\sin \theta \right) \\\\ = \left( \frac{\sqrt {3} - 1}{2}+\frac{1}{\sqrt {2}} \right) \left( \frac{8}{17}+\frac{15}{17} \right) \\\\

\\ =\frac{\sqrt {3} - 1+\sqrt {2}~}{2}\ast\frac{23}{17}=\frac{23}{34} \left( \sqrt {3} - 1+\sqrt {2} \right) \\\\


  

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