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Question Number 67918 by aliesam last updated on 02/Sep/19

Answered by $@ty@m123 last updated on 02/Sep/19

Let α+β=x tan x=(((1/2)+(1/3))/(1−(1/2)×(1/3)))=1⇒x=45^o ...(1) tan β=(1/3)⇒sin β=(1/(√(10))) & cos β=(3/(√(10))) ..(2) ∴ cos (12α+13β)=cos (12x+β) =cos 12xcos β−sin 12xsin β =cos 540^o ×(3/(√(10)))−sin 540^o ×(1/(√(10))) =(−1)×(3/(√(10)))−0×(3/(√(10))) =((−3)/(√(10)))

$${Let}\:\alpha+\beta={x} \\ $$$$\mathrm{tan}\:{x}=\frac{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{1}\Rightarrow{x}=\mathrm{45}^{\mathrm{o}} \:\:...\left(\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\sqrt{\mathrm{10}}}\:\&\:\mathrm{cos}\:\beta=\frac{\mathrm{3}}{\sqrt{\mathrm{10}}}\:..\left(\mathrm{2}\right) \\ $$$$\therefore\:\mathrm{cos}\:\left(\mathrm{12}\alpha+\mathrm{13}\beta\right)=\mathrm{cos}\:\left(\mathrm{12}{x}+\beta\right) \\ $$$$=\mathrm{cos}\:\mathrm{12}{x}\mathrm{cos}\:\:\beta−\mathrm{sin}\:\mathrm{12}{x}\mathrm{sin}\:\beta \\ $$$$=\mathrm{cos}\:\mathrm{540}^{\mathrm{o}} ×\frac{\mathrm{3}}{\sqrt{\mathrm{10}}}−\mathrm{sin}\:\mathrm{540}^{\mathrm{o}} ×\frac{\mathrm{1}}{\sqrt{\mathrm{10}}} \\ $$$$=\left(−\mathrm{1}\right)×\frac{\mathrm{3}}{\sqrt{\mathrm{10}}}−\mathrm{0}×\frac{\mathrm{3}}{\sqrt{\mathrm{10}}} \\ $$$$=\frac{−\mathrm{3}}{\sqrt{\mathrm{10}}} \\ $$

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