If sec displaystylealpha=frac{41}{{9}},{0}

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}}

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

Answers (1)

2020-11-08
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
\(\displaystyle{\tan},\frac{\alpha}{{2}}=\frac{{{1}- \cos{\alpha}}}{{ \sin{\alpha}}}\)
\(\displaystyle=\frac{{{1}-\frac{9}{{41}}}}{{\frac{40}{{41}}}}\)
\(\displaystyle=\frac{{\frac{32}{{41}}}}{{\frac{40}{{41}}}}\)
\(\displaystyle=\frac{32}{{40}}\)
\(\displaystyle{\tan},\frac{\alpha}{{2}}=\frac{4}{{5}}\)
0

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