Question

If $\sin 25^\circ \sin 35^\circ \sin 85^\circ = \dfrac{{\cos x^\circ }}{a}$, where argument of $\cos$ is acute and positive, then find the value of $\left( {x + a} \right)$.

Answer 1

Here, we need to find the value of the expression $\left( {x + a} \right)$. We will simplify the given equation using trigonometric identities, such that the left hand side can be compared to the right hand side expression $\dfrac{{\cos x^\circ }}{a}$, to obtain the values of $x$ and $a$. Finally, we will use the values of $x$ and $a$ to find the value of the expression $\left( {x + a} \right)$.

Formula Used: We will use the following formulas:

1. The trigonometric identities $\sin A\sin B = \dfrac{1}{2}\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]$ and $\sin A\cos B = \dfrac{1}{2}\left[ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]$.
2. The cosine of any angle $\theta$ is equal to the cosine of the negative angle $- \theta$, that is $\cos \theta = \cos \left( { - \theta } \right)$.
3. The sine and cosine of complementary angles is given by $\sin \left( {90^\circ - \theta } \right) = \cos \theta$.
4. The sine of an acute angle $\theta$ can be written as $\sin \left( {90^\circ + \theta } \right) = \cos \theta$, because sine is positive in the second quadrant.

Complete step by step solution:
We will use trigonometric identities to simplify the left hand side of the given equation.
Using the trigonometric identity $\sin A\sin B = \dfrac{1}{2}\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]$ to rewrite the expression $\sin 25^\circ \sin 35^\circ$, we get
$\begin{array}{l} \Rightarrow \sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\left[ {\cos \left( {25^\circ - 35^\circ } \right) - \cos \left( {25^\circ + 35^\circ } \right)} \right]\\\ \Rightarrow \sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\left[ {\cos \left( { - 10^\circ } \right) - \cos \left( {60^\circ } \right)} \right]\end{array}$
Using the identity, $\cos \theta = \cos \left( { - \theta } \right)$, we get
$\cos \left( { - 10^\circ } \right) = \cos 10^\circ$
Substituting $\cos \left( { - 10^\circ } \right) = \cos 10^\circ$ in the equation $\sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\left[ {\cos \left( { - 10^\circ } \right) - \cos \left( {60^\circ } \right)} \right]$, we get
$\Rightarrow \sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\left[ {\cos 10^\circ - \cos 60^\circ } \right]$
The cosine of the angle measuring $60^\circ$ is equal to $\dfrac{1}{2}$.
Substituting $\cos 60^\circ = \dfrac{1}{2}$ in the equation $\sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\left[ {\cos 10^\circ - \cos 60^\circ } \right]$, we get
$\Rightarrow \sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\left[ {\cos 10^\circ - \dfrac{1}{2}} \right]$
Simplifying the expression using the distributive law of multiplication, we get
$\Rightarrow \sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\cos 10^\circ - \dfrac{1}{4}$
Now, substituting $\sin 25^\circ \sin 35^\circ = \dfrac{1}{2}\cos 10^\circ - \dfrac{1}{4}$ in the equation $\sin 25^\circ \sin 35^\circ \sin 85^\circ = \dfrac{{\cos x^\circ }}{a}$, we get
$\begin{array}{l} \Rightarrow \left( {\dfrac{1}{2}\cos 10^\circ - \dfrac{1}{4}} \right)\sin 85^\circ = \dfrac{{\cos x^\circ }}{a}\\\ \Rightarrow \dfrac{1}{2}\cos 10^\circ \sin 85^\circ - \dfrac{1}{4}\sin 85^\circ = \dfrac{{\cos x^\circ }}{a}\end{array}$
We know that $\sin A\cos B = \dfrac{1}{2}\left[ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]$.
Using the trigonometric identity $\sin A\cos B = \dfrac{1}{2}\left[ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]$ to rewrite the expression $\cos 10^\circ \sin 85^\circ$, we get
$\begin{array}{l} \Rightarrow \sin 85^\circ \cos 10^\circ = \dfrac{1}{2}\left[ {\sin \left( {85^\circ + 10^\circ } \right) + \sin \left( {85^\circ - 10^\circ } \right)} \right]\\\ \Rightarrow \sin 85^\circ \cos 10^\circ = \dfrac{1}{2}\left[ {\sin 95^\circ + \sin 75^\circ } \right]\end{array}$
Simplifying the expression, we get
$\Rightarrow \cos 10^\circ \sin 85^\circ = \dfrac{1}{2}\sin 95^\circ + \dfrac{1}{2}\sin 75^\circ$
Substituting $\cos 10^\circ \sin 85^\circ = \dfrac{1}{2}\sin 95^\circ + \dfrac{1}{2}\sin 75^\circ$ in the equation $\dfrac{1}{2}\cos 10^\circ \sin 85^\circ - \dfrac{1}{4}\sin 85^\circ = \dfrac{{\cos x^\circ }}{a}$, we get
$\Rightarrow \dfrac{1}{2}\left( {\dfrac{1}{2}\sin 95^\circ + \dfrac{1}{2}\sin 75^\circ } \right) - \dfrac{1}{4}\sin 85^\circ = \dfrac{{\cos x^\circ }}{a}$
Multiplying the terms, we get
$\Rightarrow \dfrac{1}{4}\sin 95^\circ + \dfrac{1}{4}\sin 75^\circ - \dfrac{1}{4}\sin 85^\circ = \dfrac{{\cos x^\circ }}{a}$
Rewriting the expression, we get
$\Rightarrow \dfrac{1}{4}\sin \left( {90^\circ + 5^\circ } \right) + \dfrac{1}{4}\sin \left( {90^\circ - 15^\circ } \right) - \dfrac{1}{4}\sin \left( {90^\circ - 5^\circ } \right) = \dfrac{{\cos x^\circ }}{a}$
Now, we know that the sine and cosine of complementary angles is given by $\sin \left( {90^\circ - \theta } \right) = \cos \theta$.
The sine of an acute angle $\theta$ can be written as $\sin \left( {90^\circ + \theta } \right) = \cos \theta$, because sine is positive in the second quadrant.
Using the trigonometric identities $\sin \left( {90^\circ - \theta } \right) = \cos \theta$ and $\sin \left( {90^\circ + \theta } \right) = \cos \theta$ in the equation, we get
$\Rightarrow \dfrac{1}{4}\cos 5^\circ + \dfrac{1}{4}\cos 15^\circ - \dfrac{1}{4}\cos 5^\circ = \dfrac{{\cos x^\circ }}{a}$
Subtracting the like terms, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{4}\cos 15^\circ = \dfrac{{\cos x^\circ }}{a}\\\ \Rightarrow \dfrac{{\cos 15^\circ }}{4} = \dfrac{{\cos x^\circ }}{a}\end{array}$
Comparing the terms on both the sides of the equation, we get
$x = 15$ and $4 = a$
Now, we can find the value of the expression $\left( {x + a} \right)$.
Substituting $x = 15$ and $a = 4$ in the expression, we get
$x + a = 15 + 4 = 19$
Therefore, we get the value of the expression $\left( {x + a} \right)$ as 19.

Note:
We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$.
The identity $\sin A\sin B = \dfrac{1}{2}\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]$ is obtained by rewriting the trigonometric identity $\cos \left( {A + B} \right) - \cos \left( {A - B} \right) = - 2\sin A\sin B$.
The identity $\sin A\cos B = \dfrac{1}{2}\left[ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]$ is obtained by rewriting the trigonometric identity $\sin \left( {A + B} \right) + \sin \left( {A - B} \right) = 2\sin A\cos B$.