Question 262433
here is our equation
{{{f(theta)=3tan(theta)+sin^2(theta)}}}
find f(pi/6)
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step 1 - replace theta with pi/6 and we get
{{{f(pi/6) = 3tan(pi/6) + sin^2(pi/6)}}}
step 2 - find the values of
3tan(pi/6) and sin^2(pi/6) as
{{{3tan(pi/6)}}} = 3*(sqrt(3)/3) = {{{sqrt(3)}}}
{{{sin^2(pi/6)}}} = (1/2)^2 = {{{1/4}}}
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So, we get
{{{f(pi/6) = 3tan(pi/6) + sin^2(pi/6)}}}
which becomes
{{{f(pi/6) = sqrt(3) + 1/4}}}
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If however you meant the original problem as:
{{{f(theta)=3tan(theta)+sin(2(theta))}}}
then we have
step 1 - replace theta with pi/6 and we get
{{{f(pi/6) = 3tan(pi/6) + sin(2(pi/6))}}}
step 2 - find the values of
3tan(pi/6) and sin(2(pi/6)) as
{{{3tan(pi/6)}}} = 3*(sqrt(3)/3) = {{{sqrt(3)}}}
{{{sin(2(pi/6))}}} = {{{sin(pi/3)}}} = {{{sqrt(3)/2}}}
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So, we get
{{{f(pi/6) = 3tan(pi/6) + sin(2(pi/6))}}}
which becomes
{{{f(pi/6) = sqrt(3) + sqrt(3)/2}}}
or
{{{f(pi/6) = 3sqrt(3)/2}}}