Question Number 227368 by Math1 last updated on 18/Jan/26
$$\mathrm{If}\:\:\:\mathrm{sin9}°\:=\:\mathrm{a} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{cos3}°}{\mathrm{sin15}°}\:−\:\frac{\mathrm{sin3}°}{\mathrm{cos15}°}\:=\:? \\ $$
Answered by Kassista last updated on 18/Jan/26
$${Consider}\:{the}\:{arguments}\:{in}\:{degrees} \\ $$$$\frac{{cos}\:\left(\mathrm{3}\right)}{{sin}\:\left(\mathrm{15}\right)}−\frac{{sin}\left(\mathrm{3}\right)}{{cos}\left(\mathrm{15}\right)}\:=\:\frac{{cos}\left(\mathrm{3}\right){cos}\left(\mathrm{15}\right)−{sin}\left(\mathrm{3}\right){sin}\left(\mathrm{15}\right)}{{sin}\left(\mathrm{15}\right){cos}\left(\mathrm{15}\right)} \\ $$$$ \\ $$$$=\:\frac{{cos}\left(\mathrm{3}+\mathrm{15}\right)}{\frac{{sin}\left(\mathrm{30}\right)}{\mathrm{2}}}=\frac{\mathrm{2}{cos}\left(\mathrm{18}\right)}{\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\mathrm{4}{cos}\left(\mathrm{18}\right)\:=\:\mathrm{4}{cos}\left(\mathrm{2}×\mathrm{9}\right) \\ $$$$=\mathrm{4}\left(\:\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{9}\right)\:\right)=\:\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}^{\mathrm{2}} \right) \\ $$$$ \\ $$
Answered by A5T last updated on 19/Jan/26
$$?=\frac{\mathrm{cos3}°\mathrm{cos15}°−\mathrm{sin3}°\mathrm{sin15}°}{\mathrm{sin15}°\mathrm{cos15}°}=\frac{\mathrm{cos}\left(\mathrm{3}+\mathrm{15}\right)}{\frac{\mathrm{sin30}}{\mathrm{2}}} \\ $$$$\Rightarrow?=\frac{\mathrm{cos}\left(\mathrm{9}+\mathrm{9}\right)}{\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{9}−\mathrm{sin}^{\mathrm{2}} \mathrm{9}}{\frac{\mathrm{1}}{\mathrm{4}}}=\mathrm{4}\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{9}\right) \\ $$$$=\mathrm{4}−\mathrm{8a}^{\mathrm{2}} \\ $$