Question Number 83229 by peter frank last updated on 28/Feb/20
Commented by jagoll last updated on 28/Feb/20
$$\left(\mathrm{iii}\right)\:\frac{\mathrm{cos}\:\mathrm{3A}−\mathrm{cos}\:\mathrm{9A}+\mathrm{cos}\:\mathrm{A}−\mathrm{cos}\:\mathrm{3A}}{\mathrm{sin}\:\mathrm{9A}−\mathrm{sin}\:\mathrm{3A}+\mathrm{sin}\:\mathrm{3A}−\mathrm{sin}\:\mathrm{A}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{A}−\mathrm{cos}\:\mathrm{9A}}{\mathrm{sin}\:\mathrm{9A}−\mathrm{sin}\:\mathrm{A}}\:=\:\frac{\mathrm{2sin}\:\mathrm{5Asin}\:\mathrm{4A}}{\mathrm{2cos}\:\mathrm{5Asin}\:\mathrm{4A}} \\ $$$$=\:\mathrm{tan}\:\mathrm{5A} \\ $$
Commented by peter frank last updated on 28/Feb/20
$${thank}\:{you}\:{sir}. \\ $$$${is}\:{it}\:{possible}\:{to}\:{find}\: \\ $$$${general}\:{value}\:{in}\:\:{b}\:\left({i}\right)? \\ $$
Commented by jagoll last updated on 28/Feb/20
$$\left(\mathrm{i}\right)\:\mathrm{cos}\:\mathrm{px}+\:\mathrm{cos}\:\mathrm{qx}\:=\mathrm{0} \\ $$$$\mathrm{2cos}\:\left(\frac{\mathrm{p}+\mathrm{q}}{\mathrm{2}}\right)\mathrm{x}\:\mathrm{cos}\:\left(\frac{\mathrm{p}−\mathrm{q}}{\mathrm{2}}\right)\mathrm{x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left(\frac{\mathrm{p}+\mathrm{q}}{\mathrm{2}}\right)\mathrm{x}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{p}+\mathrm{q}}{\mathrm{2}}\right)\mathrm{x}\:=\:\pm\:\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi\:\Rightarrow\:\mathrm{x}\:=\:\frac{\:\pm\:\pi+\mathrm{4k}\pi}{\mathrm{p}+\mathrm{q}} \\ $$$$\Rightarrow\mathrm{similarly}\:\mathrm{cos}\:\left(\frac{\mathrm{p}−\mathrm{q}}{\mathrm{2}}\right)\mathrm{x}\:=\:\pm\:\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi \\ $$$$\mathrm{x}\:=\:\frac{\pm\:\pi+\mathrm{4k}\pi}{\mathrm{p}−\mathrm{q}} \\ $$
Commented by peter frank last updated on 28/Feb/20
$${thank}\:{you} \\ $$