Question

If \[\dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \cos x\], then find the value of \[ x = \]
A. \[4\theta \]
B. \[2\theta \]
C. \[\theta \]
D. \[3\theta \]

Answer 1

Hint: We solve the fraction in the left hand side of the equation and write the term in function of cosine. Use the expansion of \[\sin 3\theta \] and \[\sin 2\theta \] to expand the terms in the fraction. Cancel common terms from fraction and again take common terms from possible terms in the fraction and solve using the identity \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]. Use the formula \[(a - b)(a + b) = {a^2} - {b^2}\] to break and group the terms in the numerator. Apply inverse trigonometric of the same function and calculate the value of ‘x’.
* \[\sin 2\theta = 2\sin \theta \cos \theta \]
* \[\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \]
*\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]
*\[(a - b)(a + b) = {a^2} - {b^2}\]

Complete step-by-step solution:
We are given that \[\dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \cos x\]................… (1)
We solve the left hand side of the equation
\[ \Rightarrow \]LHS\[ = \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }}\]
We can write \[{\sin ^2}(2\theta ) = {\left( {\sin 2\theta } \right)^2}\]
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\sin 3\theta - \sin \theta {{(\sin 2\theta )}^2}}}{{\sin \theta + \sin 2\theta \cos \theta }}\]
Use the substitution of \[\sin 2\theta = 2\sin \theta \cos \theta \] in both numerator and denominator of the given fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\sin 3\theta - \sin \theta {{(2\sin \theta \cos \theta )}^2}}}{{\sin \theta + \left( {2\sin \theta \cos \theta } \right)\cos \theta }}\]
Square the terms inside the bracket in the numerator of the given fraction
 \[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\sin 3\theta - \sin \theta (4{{\sin }^2}\theta {{\cos }^2}\theta )}}{{\sin \theta + \left( {2\sin \theta \cos \theta } \right)\cos \theta }}\]
Multiply the terms inside the bracket with the term outside the bracket in both numerator and denominator of the given fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\sin 3\theta - 4{{\sin }^3}\theta {{\cos }^2}\theta }}{{\sin \theta + 2\sin \theta {{\cos }^2}\theta }}\]
Substitute the value of \[\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \] in the numerator of the fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{3\sin \theta - 4{{\sin }^3}\theta - 4{{\sin }^3}\theta {{\cos }^2}\theta }}{{\sin \theta + 2\sin \theta {{\cos }^2}\theta }}\]
Take \[\sin \theta \] common from all the terms in both numerator and denominator of the fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\sin \theta \left( {3 - 4{{\sin }^2}\theta - 4{{\sin }^2}\theta {{\cos }^2}\theta } \right)}}{{\sin \theta \left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Cancel the common factor i.e.\[\sin \theta \] from both numerator and denominator of the given fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\left( {3 - 4{{\sin }^2}\theta - 4{{\sin }^2}\theta {{\cos }^2}\theta } \right)}}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Now we take\[ - 4{\sin ^2}\theta \]common from the two terms in numerator of the fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{3 - 4{{\sin }^2}\theta \left( {1 + {{\cos }^2}\theta } \right)}}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Substitute the value of \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]in numerator of the fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{3 - 4\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right)}}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Since we know the identity\[(a - b)(a + b) = {a^2} - {b^2}\], we can write \[\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right) = 1 - {({\cos ^2}\theta )^2}\]i.e. \[\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right) = 1 - {\cos ^4}\theta \]
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{3 - 4\left( {1 - {{\cos }^4}\theta } \right)}}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Multiply the terms inside the bracket with the term outside the bracket in numerator of the given fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{3 - 4 + 4{{\cos }^4}\theta }}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{4{{\cos }^4}\theta - 1}}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Using the identity \[(a - b)(a + b) = {a^2} - {b^2}\] we can write \[4{\cos ^4}\theta - 1 = \left( {2{{\cos }^2}\theta - 1} \right)\left( {2{{\cos }^2}\theta + 1} \right)\]
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \dfrac{{\left( {2{{\cos }^2}\theta - 1} \right)\left( {2{{\cos }^2}\theta + 1} \right)}}{{\left( {1 + 2{{\cos }^2}\theta } \right)}}\]
Cancel the common factor i.e. \[\left( {1 + 2{{\cos }^2}\theta } \right)\] from both numerator and denominator of the given fraction
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \left( {2{{\cos }^2}\theta - 1} \right)\]
Use the identity \[\left( {2{{\cos }^2}\theta - 1} \right) = \cos 2\theta \] in RHS
\[ \Rightarrow \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \cos 2\theta \]...............… (2)
Since we are given from equation (1)\[\dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }} = \cos x\]
Equate the values of fraction from equations (1) and (2)
\[ \Rightarrow \cos x = \cos 2\theta \]
Apply inverse cosine function on both sides of the equation
\[ \Rightarrow {\cos ^{ - 1}}\left( {\cos x} \right) = {\cos ^{ - 1}}\left( {\cos 2\theta } \right)\]
Cancel inverse function by the same function
\[ \Rightarrow x = 2\theta \]
\[\therefore \]The value of x is \[2\theta \]

\[\therefore \]Option B is correct.

Note: Many students get confused while grouping the factors\[\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right)\]and write their multiplication using the identity as \[\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right) = 1 - {\left( {{{\cos }^2}\theta } \right)^2}\]. Keep in mind we can write \[{\cos ^2}\theta = {\left( {\cos \theta } \right)^2}\]and we know \[{\left( {{m^p}} \right)^q} = {m^{p \times q}}\]so the value of\[{\left( {{{\cos }^2}\theta } \right)^2} = {\left( {{{\left( {\cos \theta } \right)}^2}} \right)^2} = {\cos ^4}\theta \]. Also, many students write the fraction in terms of sine function which is wrong as then we will not be able to apply the inverse of cosine on both sides.