Hi, need some help on the following question!! find the possible values of tan(1/2)x, if tanx=3/4 thanks A Ui need the method explained as clearly...

Thursday, February 5, 2015

$\displaystyle{\tan{{\left({x}\right)}}}=\frac{{3}}{{4}}$

We need to
find the value of $\displaystyle{\tan{{\left(\frac{{1}}{{2}}\right)}}}{x}={\tan{{\left(\frac{{x}}{{2}}\right)}}}$

$\displaystyle{W}{e}{k}{n}{o}{w}$
that:

$\displaystyle{\tan{{\left({x}\right)}}}=$
sin(x)/cos(x)

$\displaystyle=\Rightarrow\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}=$
3/4

$\displaystyle=\Rightarrow{3}{\cos{{\left({x}\right)}}}=$
4sin(x)

Now we know that $\displaystyle{{\sin}}^{{2}}{x}+{{\cos}}^{{2}}{x}={1}=\Rightarrow$
cosx = sqrt(1-sin^2 x)

$\displaystyle=\Rightarrow{3}\sqrt{{{1}-{{\sin}}^{{2}}{x}}}=$
4sin(x)

Now we will square both
sides:

$\displaystyle=\Rightarrow{9}{\left({1}-{{\sin}}^{{2}}{x}\right)}={14}{{\sin}}^{{2}}$
x

$\displaystyle=\Rightarrow{9}-{9}{{\sin}}^{{2}}{x}=={14}{{\sin}}^{{2}}$
x

$\displaystyle=\Rightarrow{25}{{\sin}}^{{2}}{x}=$
9

$\displaystyle=\Rightarrow{{\sin}}^{{2}}{x}=$
9/25

$\displaystyle=\Rightarrow{\sin{{\left({x}\right)}}}=$
3/5

$\displaystyle=\Rightarrow{\cos{{\left({x}\right)}}}=$
4/5 $\displaystyle\ldots\ldots\ldots\ldots\ldots\ldots{\left({2}\right)}$

Now we know
that:

$\displaystyle{\cos{{\left({2}{x}\right)}}}={1}-{2}{{\sin}}^{{2}}$
x

$\displaystyle=\Rightarrow{\cos{{\left({x}\right)}}}={1}-{2}{{\sin}}^{{2}}$
(x/2)

$\displaystyle=\Rightarrow\frac{{4}}{{5}}={1}-{2}{{\sin}}^{{2}}$
(x/2)

$\displaystyle=\Rightarrow{2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}={1}-$
4/5

$\displaystyle=\Rightarrow{2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}=$
1/5

$\displaystyle=\Rightarrow{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}=$
1/10

$\displaystyle=\Rightarrow{\sin{{\left(\frac{{x}}{{2}}\right)}}}=$
1/sqrt10

Now we know
that:

$\displaystyle{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}+{{\cos}}^{{2}}{\left(\frac{{x}}{{2}}\right)}=$
1

$\displaystyle=\Rightarrow{\cos{{\left(\frac{{x}}{{2}}\right)}}}=\sqrt{{{1}-{{\sin}}^{{2}}}}$
(x/2))

$\displaystyle=\Rightarrow{\cos{{\left(\frac{{x}}{{2}}\right)}}}=\sqrt{{{1}-}}$
1/10)

$\displaystyle=\Rightarrow{\cos{{\left(\frac{{x}}{{2}}\right)}}}=\sqrt{{\frac{{9}}{{10}}}}=$
3/sqrt10

$\displaystyle=\Rightarrow{\tan{{\left(\frac{{x}}{{2}}\right)}}}=\frac{{\sin{{\left(\frac{{x}}{{2}}\right)}}}}{}$
cos(x/2)

$\displaystyle=\Rightarrow{\tan{{\left(\frac{{x}}{{2}}\right)}}}=\frac{{\frac{{1}}{\sqrt{{10}}}}}{{\frac{{3}}{\sqrt{{10}}}}}=$
1/3

$\displaystyle=\Rightarrow{\tan{{\left(\frac{{x}}{{2}}\right)}}}=$
1/3

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