Question Number 31188 by mondodotto@gmail.com last updated on 03/Mar/18
Commented by Tinkutara last updated on 03/Mar/18
![tan θ=(4/3) sin θ=±(4/5);cos θ=±(3/5) ±(3/5)=1−2sin^2 (θ/2) sin^2 (θ/2)=(1/5) sin (θ/2)=(1/( (√5))) [∵0≤(θ/2)≤π] cos (θ/2)=±(2/( (√5))) tan (θ/2)=±(1/2)](https://www.tinkutara.com/question/Q31192.png)
$$\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{sin}\:\theta=\pm\frac{\mathrm{4}}{\mathrm{5}};\mathrm{cos}\:\theta=\pm\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\pm\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\left[\because\mathrm{0}\leqslant\frac{\theta}{\mathrm{2}}\leqslant\pi\right] \\ $$$$\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mondodotto@gmail.com last updated on 03/Mar/18
$$\mathrm{i}\:\mathrm{still}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{you}\:\mathrm{sir} \\ $$