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Question: If \[\dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \th...

If sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=cos⁡x\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=x =
A. 4θ4\theta
B. 2θ2\theta
C. θ\theta
D. 3θ3\theta

Explanation

Solution

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⁡3θ\sin 3\theta and sin⁡2θ\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⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta . Use the formula (a−b)(a+b)=a2−b2(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⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
  • sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta
    *sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta
    *(a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}

Complete step-by-step solution:
We are given that sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=cos⁡x\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
⇒\RightarrowLHS=sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ = \dfrac{{\sin 3\theta - \sin \theta {{\sin }^2}(2\theta )}}{{\sin \theta + \sin 2\theta \cos \theta }}
We can write sin⁡2(2θ)=(sin⁡2θ)2{\sin ^2}(2\theta ) = {\left( {\sin 2\theta } \right)^2}
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=sin⁡3θ−sin⁡θ(sin⁡2θ)2sin⁡θ+sin⁡2θcos⁡θ\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⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta in both numerator and denominator of the given fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=sin⁡3θ−sin⁡θ(2sin⁡θcos⁡θ)2sin⁡θ+(2sin⁡θcos⁡θ)cos⁡θ\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
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=sin⁡3θ−sin⁡θ(4sin⁡2θcos⁡2θ)sin⁡θ+(2sin⁡θcos⁡θ)cos⁡θ\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
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=sin⁡3θ−4sin⁡3θcos⁡2θsin⁡θ+2sin⁡θcos⁡2θ\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⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta in the numerator of the fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=3sin⁡θ−4sin⁡3θ−4sin⁡3θcos⁡2θsin⁡θ+2sin⁡θcos⁡2θ\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⁡θ\sin \theta common from all the terms in both numerator and denominator of the fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=sin⁡θ(3−4sin⁡2θ−4sin⁡2θcos⁡2θ)sin⁡θ(1+2cos⁡2θ)\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⁡θ\sin \theta from both numerator and denominator of the given fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=(3−4sin⁡2θ−4sin⁡2θcos⁡2θ)(1+2cos⁡2θ)\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−4sin⁡2θ- 4{\sin ^2}\thetacommon from the two terms in numerator of the fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=3−4sin⁡2θ(1+cos⁡2θ)(1+2cos⁡2θ)\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⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta in numerator of the fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=3−4(1−cos⁡2θ)(1+cos⁡2θ)(1+2cos⁡2θ)\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)=a2−b2(a - b)(a + b) = {a^2} - {b^2}, we can write (1−cos⁡2θ)(1+cos⁡2θ)=1−(cos⁡2θ)2\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right) = 1 - {({\cos ^2}\theta )^2}i.e. (1−cos⁡2θ)(1+cos⁡2θ)=1−cos⁡4θ\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right) = 1 - {\cos ^4}\theta
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=3−4(1−cos⁡4θ)(1+2cos⁡2θ)\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
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=3−4+4cos⁡4θ(1+2cos⁡2θ)\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)}}
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=4cos⁡4θ−1(1+2cos⁡2θ)\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)=a2−b2(a - b)(a + b) = {a^2} - {b^2} we can write 4cos⁡4θ−1=(2cos⁡2θ−1)(2cos⁡2θ+1)4{\cos ^4}\theta - 1 = \left( {2{{\cos }^2}\theta - 1} \right)\left( {2{{\cos }^2}\theta + 1} \right)
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=(2cos⁡2θ−1)(2cos⁡2θ+1)(1+2cos⁡2θ)\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. (1+2cos⁡2θ)\left( {1 + 2{{\cos }^2}\theta } \right) from both numerator and denominator of the given fraction
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=(2cos⁡2θ−1)\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 (2cos⁡2θ−1)=cos⁡2θ\left( {2{{\cos }^2}\theta - 1} \right) = \cos 2\theta in RHS
⇒sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=cos⁡2θ\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)sin⁡3θ−sin⁡θsin⁡2(2θ)sin⁡θ+sin⁡2θcos⁡θ=cos⁡x\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)
⇒cos⁡x=cos⁡2θ\Rightarrow \cos x = \cos 2\theta
Apply inverse cosine function on both sides of the equation
⇒cos⁡−1(cos⁡x)=cos⁡−1(cos⁡2θ)\Rightarrow {\cos ^{ - 1}}\left( {\cos x} \right) = {\cos ^{ - 1}}\left( {\cos 2\theta } \right)
Cancel inverse function by the same function
⇒x=2θ\Rightarrow x = 2\theta
∴\therefore The value of x is 2θ2\theta

∴\therefore Option B is correct.

Note: Many students get confused while grouping the factors(1−cos⁡2θ)(1+cos⁡2θ)\left( {1 - {{\cos }^2}\theta } \right)\left( {1 + {{\cos }^2}\theta } \right)and write their multiplication using the identity as (1−cos⁡2θ)(1+cos⁡2θ)=1−(cos⁡2θ)2\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⁡2θ=(cos⁡θ)2{\cos ^2}\theta = {\left( {\cos \theta } \right)^2}and we know (mp)q=mp×q{\left( {{m^p}} \right)^q} = {m^{p \times q}}so the value of(cos⁡2θ)2=((cos⁡θ)2)2=cos⁡4θ{\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.