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

Hint:
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 \[\sin \left( {A + B} \right) + \sin \left( {A - B} \right) = 2\sin A\cos B\].