Question

Find the value of x \[\dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \tan xA\].

Answer 1

Hint: Here, we will simplify the left-hand side using various properties of trigonometric functions. We will convert the left-hand side into a function of tangent. Then we will compare the Left-hand side and the Right-hand side and find the value of \[x\].

Formulas used: We will use following formulas:
1.\[3\sin A = 3\sin A - 4{\sin ^3}A\]
2.\[\cos 3A = 4{\cos ^3}A - 3\cos A\]
3.\[\sin 2A = 2\sin A\cos A\]
4.\[\cos 2A = 2{\cos ^2}A - 1\]
5.\[{\sin ^2}A + {\cos ^2}A = 1\]

Complete step-by-step answer:
We will start by simplifying the numerator. We know that \[\sin 3A\] is the same as .., we will substitute the value in the numerator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{\sin A + \left( {3\sin A - 4{{\sin }^3}A} \right)}}{{\cos A + \cos 3A}}\]
We know that \[\cos 3A\] is the same as \[4{\cos ^3}A - 3\cos A\]. We will substitute the value in the denominator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{\sin A + 3\sin A - 4{{\sin }^3}A}}{{\cos A + 4{{\cos }^3}A - 3\cos A}}\]
We will add the like terms:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{4\sin A - 4{{\sin }^3}A}}{{4{{\cos }^3}A - 2\cos A}}\]
We will take out the common factors of terms in the numerator and the denominator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{4\sin A\left( {1 - {{\sin }^2}A} \right)}}{{2\cos A\left( {2{{\cos }^2}A - 1} \right)}}\]
We know that:
\[\begin{array}{l} \Rightarrow {\cos ^2}x + {\sin ^2}A = 1\\ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}A\end{array}\]
We will substitute \[{\cos ^2}A\] for \[1 - {\sin ^2}A\] in the numerator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{4\sin A{{\cos }^2}A}}{{2\cos A\left( {2{{\cos }^2}A - 1} \right)}}\]
We will cancel the common terms in the numerator and the denominator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{2\sin A\cos A}}{{\left( {2{{\cos }^2}A - 1} \right)}}\]
We know that \[2\sin A\cos A\] is the same as \[\sin 2A\], we will substitute the value in the numerator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{\sin 2A}}{{\left( {2{{\cos }^2}A - 1} \right)}}\]
We know that \[2{\cos ^2}A - 1\] is the same as \[\cos 2A\]. We will substitute the value in the denominator:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \dfrac{{\sin 2A}}{{\cos 2A}}\]
We will simplify the above expression using the formula for the tangent of an angle; that is, the ratio of the Sine of an angle to the Cosine of an angle is equal to the Tangent of that angle:
\[ \Rightarrow \dfrac{{\sin A + \sin 3A}}{{\cos A + \cos 3A}} = \tan 2A\]
We will compare the Left-hand side with the Right-hand side:
\[\begin{array}{l}\tan 2A = \tan xA\\ \Rightarrow 2 = x\end{array}\]
$\therefore $ The value of \[x\] is 2.

Note: We know that that the Sine and Cosine of the sum of 2 angles is:
\[\begin{array}{l}\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\\\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\end{array}\]
We can derive the formulas of \[\sin 2A,\sin 3A,\cos 2A\] and \[\cos 3A\] that we have used in the solution using the above formulas.