# If sec displaystylealpha=frac{41}{{9}},{0}<alpha<frac{pi}{{2}}, then find the exact value of each of the following.a) displaystyle{sin},frac{alpha}{{2}}b) displaystyle{cos},frac{alpha}{{2}}c) displaystyle{tan},frac{alpha}{{2}}

If sec $$\displaystyle\alpha=\frac{41}{{9}},{0}<\alpha<\frac{\pi}{{2}},$$ then find the exact value of each of the following.
a) $$\displaystyle{\sin},\frac{\alpha}{{2}}$$
b) $$\displaystyle{\cos},\frac{\alpha}{{2}}$$
c) $$\displaystyle{\tan},\frac{\alpha}{{2}}$$

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cyhuddwyr9
Step 1
Given that,
$$\displaystyle \sec{\alpha}=\frac{41}{{9}}\ \text{and}\ \alpha$$ is in furst quadrant.
In first quandrant, all the trigonometric angles are positive.
$$\displaystyle \sec{\alpha}=\frac{{{h}{y}{p}}}{{{a}{d}{j}}}=\frac{41}{{9}}$$
Using Pythagorean Theorem,
$$\displaystyle{\left({o}{p}{p}\right)}^{2}={\left({h}{y}{p}\right)}^{2}-{\left({a}{d}{j}\right)}^{2}$$
$$\displaystyle={\left({41}\right)}^{2}-{\left({9}\right)}^{2}$$
$$= 1681\ -\ 81$$
$$= 1600$$
$$\displaystyle{o}{p}{p}=\pm\sqrt{{1600}}$$
$$\displaystyle{o}{p}{p}=\pm{40}$$
Since, all the angles area positive we take $$opp = 40$$
Step 2
a) $$\displaystyle{\sin},\frac{\alpha}{{2}}:$$
Using the formula,
$$\displaystyle{\sin},\frac{\alpha}{{2}}=\frac{\sqrt{{{1}- \cos{\alpha}}}}{{2}}$$ ... (1)
$$\displaystyle \cos{\alpha}=\frac{1}{{ \sec{\alpha}}}=\frac{1}{{\frac{41}{{9}}}}=\frac{9}{{41}}$$
Step 3
Substitute the above value in equation (1), we get
$$\displaystyle{\sin},\frac{\alpha}{{2}}=\sqrt{{\frac{{{1}- \cos{\alpha}}}{{2}}}}$$
$$\displaystyle{\sin},\frac{\alpha}{{2}}=\sqrt{{\frac{{{1}-\frac{9}{{41}}}}{{2}}}}$$
$$\displaystyle=\sqrt{{\frac{{\frac{{{41}-{9}}}{{41}}}}{{2}}}}$$
$$\displaystyle=\sqrt{{\frac{32}{{{41}\times{2}}}}}$$
$$\displaystyle=\sqrt{{\frac{16}{{41}}}}$$
$$\displaystyle=\frac{\sqrt{{16}}}{\sqrt{{41}}}$$
$$\displaystyle=\frac{4}{\sqrt{{41}}}$$
$$\displaystyle=\frac{4}{\sqrt{{41}}}\times\frac{\sqrt{{41}}}{\sqrt{{41}}}$$
$$\displaystyle{\sin},\frac{\alpha}{{2}}=\frac{{{4}\sqrt{{41}}}}{{41}}$$
Step 4
b) $$\displaystyle{\cos},\frac{\alpha}{{2}}$$
Using the formula,
$$\displaystyle{\cos},\frac{\alpha}{{2}}=\sqrt{{\frac{{{1}+ \cos{\alpha}}}{{2}}}}$$..(2)
Substitute $$\displaystyle \cos{\alpha}=\frac{9}{{41}}$$ in the above formula, we get
$$\displaystyle{\cos},\frac{\alpha}{{2}}=\sqrt{{\frac{{{1}+ \cos{\alpha}}}{{2}}}}$$
$$\displaystyle=\sqrt{{\frac{{{1}+\frac{9}{{41}}}}{{2}}}}$$
$$\displaystyle=\sqrt{{\frac{50}{{{41}\times{2}}}}}$$
$$\displaystyle=\sqrt{{\frac{25}{{41}}}}$$
$$\displaystyle=\frac{\sqrt{{25}}}{\sqrt{{41}}}$$
$$\displaystyle=\frac{5}{\sqrt{{41}}}$$
$$\displaystyle=\frac{5}{\sqrt{{41}}}\times\frac{\sqrt{{41}}}{\sqrt{{41}}}$$
$$\displaystyle{\cos},\frac{\alpha}{{2}}=\frac{{{5}\sqrt{{41}}}}{{41}}$$
Step 5
c) $$\displaystyle{\tan},\frac{\alpha}{{2}}$$
Using the formula,
$$\displaystyle{\tan},\frac{\alpha}{{2}}=\frac{{{1}- \cos{\alpha}}}{{ \sin{\alpha}}}$$...(3)
$$\displaystyle \sin{\alpha}=\frac{{{o}{p}{p}}}{{{h}{y}{p}}}=\frac{40}{{41}}$$
$$\displaystyle \cos{\alpha}=\frac{9}{{41}}$$
Substitute the above values in equation (3), we get
$$\displaystyle{\tan},\frac{\alpha}{{2}}=\frac{{{1}- \cos{\alpha}}}{{ \sin{\alpha}}}$$
$$\displaystyle=\frac{{{1}-\frac{9}{{41}}}}{{\frac{40}{{41}}}}$$
$$\displaystyle=\frac{{\frac{32}{{41}}}}{{\frac{40}{{41}}}}$$
$$\displaystyle=\frac{32}{{40}}$$
$$\displaystyle{\tan},\frac{\alpha}{{2}}=\frac{4}{{5}}$$