How can I solve this sin, cos, tan mathematical equation?
$\sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$
Answer
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Hint: The above trigonometric expression can be solved in 3 steps. Let A = ${{\cos }^{-1}}\left( \dfrac{63}{65} \right)$;
B = $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$.In step1, we convert the function A into an inverse tangent function. In step2, we convert the function B into an inverse tangent function. In step3, we simplify the functions in step1 further to get the expression in the form $\sin \left( {{\sin }^{-1}}\theta \right)$.
Complete step by step solution:
The given trigonometric equation is
$\sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$
Let A = ${{\cos }^{-1}}\left( \dfrac{63}{65} \right)$;
B = $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$
Step1:
We need to convert the function A into inverse tangent function.
From the formulae of trigonometry,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{a}{b} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{a} \right)$
Applying the same formula for ${{\cos }^{-1}}\left( \dfrac{63}{65} \right)$,
we get,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{{{65}^{2}}-{{63}^{2}}}}{63} \right)$
Writing the values of squares,
we get,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{4225-3969}}{63} \right)$
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{256}}{63} \right)$
We know that $\sqrt{256}=16$
Substituting the same,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$
Step2:
We need to convert the function B into inverse tangent function.
From the formula of trigonometry,
$\Rightarrow 2{{\tan }^{-1}}a={{\tan }^{-1}}\left( \dfrac{2a}{\left( 1-{{a}^{2}} \right)} \right)$
Applying the same formula for $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$,
we get,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{3}{4}}{1-{{\left( \dfrac{3}{4} \right)}^{2}}} \right)$
Simplifying the numerator and denominator on the right-hand side,
we get,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{2}}{1-\left( \dfrac{9}{16} \right)} \right)$
Evaluating the denominator on the right-hand side further,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{2}}{\dfrac{\left( 16-9 \right)}{16}} \right)$
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{2}}{\dfrac{7}{16}} \right)$
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{3}{2}\times \dfrac{16}{7} \right)$
Cancelling the factors,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{24}{7} \right)$
Step3:
we need to evaluate the two inverse tangent functions in step1 and step2.
The given question is of the form
$\Rightarrow \sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$
From step1 and step2,
we can write the above equation as
$\Rightarrow \sin \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)$
From the formula of trigonometry,
${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$
Applying the same formula for the above expression,
we get,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{16}{63} \right)+\left( \dfrac{24}{7} \right)}{1-\left( \dfrac{16}{63} \right)\left( \dfrac{24}{7} \right)} \right)$
Taking LCM of the numerator and denominator on right-hand side,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{\left( 16\times 7 \right)+\left( 24\times 63 \right)}{\left( 63\times 7 \right)} \right)}{\left( \dfrac{\left( 63\times 7 \right)-\left( 16\times 24 \right)}{\left( 63\times 7 \right)} \right)} \right)$
Simplifying the numerator and denominator on the right-hand side,
we get,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{112+1512}{441} \right)}{\left( \dfrac{441-384}{441} \right)} \right)$
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{1624}{441} \right)}{\left( \dfrac{54}{441} \right)} \right)$
Evaluating the above equation further,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{1624}{54} \right)$
Substituting the result in the expression
$\sin \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)$,
$\Rightarrow \sin \left( {{\tan }^{-1}}\left( \dfrac{1624}{54} \right) \right)$
Now,
From the formula of trigonometry,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{a}{b} \right)={{\sin }^{-1}}\left( \dfrac{a}{\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)} \right)$
Applying the same formula for the above expression,
we get,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{\left( \sqrt{{{\left( 1624 \right)}^{2}}+{{\left( 54 \right)}^{2}}} \right)} \right)$
Writing the values of squares on the right-hand side,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{\left( \sqrt{\left( 2637376 \right)+\left( 2916 \right)} \right)} \right)$
Evaluating the expression further,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{\left( \sqrt{\left( 2640292 \right)} \right)} \right)$
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{1625} \right)$
Substituting the above value in the expression $\sin \left( {{\tan }^{-1}}\left( \dfrac{1624}{54} \right) \right)$,
we get,
$\Rightarrow \sin \left( {{\sin }^{-1}}\left( \dfrac{1624}{1625} \right) \right)$
From the inverse trigonometry,
we know that $\sin \left( {{\sin }^{-1}}\theta \right)=\theta $
Substituting the same,
$\Rightarrow \sin \left( {{\sin }^{-1}}\left( \dfrac{1624}{1625} \right) \right)=\left( \dfrac{1624}{1625} \right)$
Therefore, the result of the above mathematical equation $\sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$ is $\left( \dfrac{1624}{1625} \right)$.
Note: We need to know the formulae of trigonometry to solve the problem easily. The above mathematical equation can also be solved by replacing the values of inverse trigonometric functions with an angle.
B = $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$.In step1, we convert the function A into an inverse tangent function. In step2, we convert the function B into an inverse tangent function. In step3, we simplify the functions in step1 further to get the expression in the form $\sin \left( {{\sin }^{-1}}\theta \right)$.
Complete step by step solution:
The given trigonometric equation is
$\sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$
Let A = ${{\cos }^{-1}}\left( \dfrac{63}{65} \right)$;
B = $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$
Step1:
We need to convert the function A into inverse tangent function.
From the formulae of trigonometry,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{a}{b} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{a} \right)$
Applying the same formula for ${{\cos }^{-1}}\left( \dfrac{63}{65} \right)$,
we get,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{{{65}^{2}}-{{63}^{2}}}}{63} \right)$
Writing the values of squares,
we get,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{4225-3969}}{63} \right)$
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{256}}{63} \right)$
We know that $\sqrt{256}=16$
Substituting the same,
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{63}{65} \right)={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$
Step2:
We need to convert the function B into inverse tangent function.
From the formula of trigonometry,
$\Rightarrow 2{{\tan }^{-1}}a={{\tan }^{-1}}\left( \dfrac{2a}{\left( 1-{{a}^{2}} \right)} \right)$
Applying the same formula for $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$,
we get,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{3}{4}}{1-{{\left( \dfrac{3}{4} \right)}^{2}}} \right)$
Simplifying the numerator and denominator on the right-hand side,
we get,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{2}}{1-\left( \dfrac{9}{16} \right)} \right)$
Evaluating the denominator on the right-hand side further,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{2}}{\dfrac{\left( 16-9 \right)}{16}} \right)$
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{2}}{\dfrac{7}{16}} \right)$
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{3}{2}\times \dfrac{16}{7} \right)$
Cancelling the factors,
$\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( \dfrac{24}{7} \right)$
Step3:
we need to evaluate the two inverse tangent functions in step1 and step2.
The given question is of the form
$\Rightarrow \sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$
From step1 and step2,
we can write the above equation as
$\Rightarrow \sin \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)$
From the formula of trigonometry,
${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$
Applying the same formula for the above expression,
we get,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{16}{63} \right)+\left( \dfrac{24}{7} \right)}{1-\left( \dfrac{16}{63} \right)\left( \dfrac{24}{7} \right)} \right)$
Taking LCM of the numerator and denominator on right-hand side,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{\left( 16\times 7 \right)+\left( 24\times 63 \right)}{\left( 63\times 7 \right)} \right)}{\left( \dfrac{\left( 63\times 7 \right)-\left( 16\times 24 \right)}{\left( 63\times 7 \right)} \right)} \right)$
Simplifying the numerator and denominator on the right-hand side,
we get,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{112+1512}{441} \right)}{\left( \dfrac{441-384}{441} \right)} \right)$
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{1624}{441} \right)}{\left( \dfrac{54}{441} \right)} \right)$
Evaluating the above equation further,
$\Rightarrow \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)={{\tan }^{-1}}\left( \dfrac{1624}{54} \right)$
Substituting the result in the expression
$\sin \left( {{\tan }^{-1}}\left( \dfrac{16}{63} \right)+{{\tan }^{-1}}\left( \dfrac{24}{7} \right) \right)$,
$\Rightarrow \sin \left( {{\tan }^{-1}}\left( \dfrac{1624}{54} \right) \right)$
Now,
From the formula of trigonometry,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{a}{b} \right)={{\sin }^{-1}}\left( \dfrac{a}{\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)} \right)$
Applying the same formula for the above expression,
we get,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{\left( \sqrt{{{\left( 1624 \right)}^{2}}+{{\left( 54 \right)}^{2}}} \right)} \right)$
Writing the values of squares on the right-hand side,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{\left( \sqrt{\left( 2637376 \right)+\left( 2916 \right)} \right)} \right)$
Evaluating the expression further,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{\left( \sqrt{\left( 2640292 \right)} \right)} \right)$
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1624}{54} \right)={{\sin }^{-1}}\left( \dfrac{1624}{1625} \right)$
Substituting the above value in the expression $\sin \left( {{\tan }^{-1}}\left( \dfrac{1624}{54} \right) \right)$,
we get,
$\Rightarrow \sin \left( {{\sin }^{-1}}\left( \dfrac{1624}{1625} \right) \right)$
From the inverse trigonometry,
we know that $\sin \left( {{\sin }^{-1}}\theta \right)=\theta $
Substituting the same,
$\Rightarrow \sin \left( {{\sin }^{-1}}\left( \dfrac{1624}{1625} \right) \right)=\left( \dfrac{1624}{1625} \right)$
Therefore, the result of the above mathematical equation $\sin \left( {{\cos }^{-1}}\left( \dfrac{63}{65} \right)+2{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$ is $\left( \dfrac{1624}{1625} \right)$.
Note: We need to know the formulae of trigonometry to solve the problem easily. The above mathematical equation can also be solved by replacing the values of inverse trigonometric functions with an angle.
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