What is the value of $\sin \left( 10 \right)\cos \left( 20 \right)\sin \left( 30 \right)\cos \left( 40 \right)\sin \left( 50 \right)\cos \left( 60 \right)\sin \left( 70 \right)\cos \left( 80 \right)$ ? Not in radians, in degrees.
Answer
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Hint: Two angles are complementary if they add up to make ${{90}^{\circ }}$ . for complementary angles, the trigonometric ratios are related to each other as
If, $\angle A+\angle B={{90}^{\circ }}$
$\begin{align}
& \sin A=\cos B \\
& \cos A=\sin B \\
\end{align}$
Certain identities and expansions will help us in this question
$\begin{align}
& \cos \left( x+y \right)=\cos x\cos y-\sin x\sin y \\
& \cos \left( x-y \right)=\cos x\cos y+\sin x\sin y \\
& \sin \left( x+y \right)=\sin x\cos y+\cos x\sin y \\
& \sin \left( x-y \right)=\sin x\cos y-\cos x\sin y \\
& {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
\end{align}$
Certain trigonometric values to be remembered:
$\begin{align}
& \sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
& \cos {{60}^{\circ }}=\sin {{30}^{\circ }}=\dfrac{1}{2} \\
& \sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\
\end{align}$
Complete step by step answer:
Here, we can see that we have pairs of complementary angles multiplied together,
Arranging them in pairs, we get
$\left( \sin \left( 10 \right)\cos \left( 80 \right) \right)\left( \cos \left( 20 \right)\sin \left( 70 \right) \right)\left( \sin \left( 30 \right)\cos \left( 60 \right) \right)\left( \cos \left( 40 \right)\sin \left( 50 \right) \right)$
Since they are complementary angles, sine of one angle is equal to the cosine of its complement. Hence, using this relation, we get
$\left( {{\sin }^{2}}\left( 10 \right) \right)\left( {{\cos }^{2}}\left( 20 \right) \right)\left( {{\sin }^{2}}\left( 30 \right) \right)\left( {{\cos }^{2}}\left( 40 \right) \right)$
Using $\sin {{30}^{\circ }}=\dfrac{1}{2}$,we have
$\dfrac{1}{4}\left( {{\sin }^{2}}\left( 10 \right) \right)\left( {{\cos }^{2}}\left( 20 \right) \right)\left( {{\cos }^{2}}\left( 40 \right) \right)$
Now, we can club ${{\cos }^{2}}\left( {{20}^{\circ }} \right)$ and ${{\cos }^{2}}\left( {{40}^{\circ }} \right)$ together and use above mentioned expansions to simplify it further
$\dfrac{1}{4}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( \cos \left( 20 \right)\cos \left( 40 \right) \right)}^{2}}$
Multiplying and dividing by 2,
$\begin{align}
& =\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( 2\cos \left( 20 \right)\cos \left( 40 \right) \right)}^{2}} \\
& =\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( \cos \left( 40+20 \right)+\cos \left( 40-20 \right) \right)}^{2}} \\
& =\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( \cos \left( 60 \right)+\cos \left( 20 \right) \right)}^{2}} \\
\end{align}$
Expanding the above expression using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$=\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right)\left( {{\cos }^{2}}\left( 60 \right)+{{\cos }^{2}}\left( 20 \right)+2\cos \left( 60 \right)\cos \left( 20 \right) \right)$
Using the value of $\cos {{60}^{\circ }}=\dfrac{1}{2}$ and including ${{\sin }^{2}}\left( 10 \right)$ inside the brackets, we get
$=\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+{{\left( \sin \left( 10 \right)\cos \left( 20 \right) \right)}^{2}}+\sin \left( 10 \right)\sin \left( 10 \right)\cos \left( 20 \right) \right)$
Now, consider $\sin \left( 10 \right)\cos \left( 20 \right)$
Multiplying and dividing by 2, we get
$\dfrac{2\sin \left( 10 \right)\cos \left( 20 \right)}{2}$
Using the above expansions, this expression will be simplified as
$\begin{align}
& =\dfrac{\sin \left( 10+20 \right)+\sin \left( 10-20 \right)}{2} \\
& =\dfrac{\sin \left( 30 \right)-\sin \left( 10 \right)}{2} \\
\end{align}$
Using the value of $\sin {{30}^{\circ }}=\dfrac{1}{2}$, we have
$=\dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2}$
Now using this value in the above formulated expression, we have
$\begin{align}
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+{{\left( \sin \left( 10 \right)\cos \left( 20 \right) \right)}^{2}}+\sin \left( 10 \right)\sin \left( 10 \right)\cos \left( 20 \right) \right) \\
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+{{\left( \dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2} \right)}^{2}}+\sin \left( 10 \right)\left( \dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2} \right) \right) \\
\end{align}$
Expanding the expression, we get
$\begin{align}
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+\left( \dfrac{1}{16}+\dfrac{{{\sin }^{2}}\left( 10 \right)}{4}-\dfrac{\sin \left( 10 \right)}{4} \right)+\sin \left( 10 \right)\left( \dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2} \right) \right) \\
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+\dfrac{1}{16}+\dfrac{{{\sin }^{2}}\left( 10 \right)}{4}-\dfrac{\sin \left( 10 \right)}{4}+\dfrac{\sin \left( 10 \right)}{4}-\dfrac{{{\sin }^{2}}\left( 10 \right)}{2} \right) \\
& =\dfrac{1}{16}\times \dfrac{1}{16} \\
& =\dfrac{1}{256} \\
\end{align}$
Hence, the final value of $\sin \left( 10 \right)\cos \left( 20 \right)\sin \left( 30 \right)\cos \left( 40 \right)\sin \left( 50 \right)\cos \left( 60 \right)\sin \left( 70 \right)\cos \left( 80 \right)$ is $\dfrac{1}{256}$ .
Note: While solving such large expressions, always keep note of the constants that are being multiplied or divided in the expression because any change in them will change the whole answer. Also, apply each trigonometric identity carefully keeping in mind the positive and negative signs.
If, $\angle A+\angle B={{90}^{\circ }}$
$\begin{align}
& \sin A=\cos B \\
& \cos A=\sin B \\
\end{align}$
Certain identities and expansions will help us in this question
$\begin{align}
& \cos \left( x+y \right)=\cos x\cos y-\sin x\sin y \\
& \cos \left( x-y \right)=\cos x\cos y+\sin x\sin y \\
& \sin \left( x+y \right)=\sin x\cos y+\cos x\sin y \\
& \sin \left( x-y \right)=\sin x\cos y-\cos x\sin y \\
& {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
\end{align}$
Certain trigonometric values to be remembered:
$\begin{align}
& \sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
& \cos {{60}^{\circ }}=\sin {{30}^{\circ }}=\dfrac{1}{2} \\
& \sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\
\end{align}$
Complete step by step answer:
Here, we can see that we have pairs of complementary angles multiplied together,
Arranging them in pairs, we get
$\left( \sin \left( 10 \right)\cos \left( 80 \right) \right)\left( \cos \left( 20 \right)\sin \left( 70 \right) \right)\left( \sin \left( 30 \right)\cos \left( 60 \right) \right)\left( \cos \left( 40 \right)\sin \left( 50 \right) \right)$
Since they are complementary angles, sine of one angle is equal to the cosine of its complement. Hence, using this relation, we get
$\left( {{\sin }^{2}}\left( 10 \right) \right)\left( {{\cos }^{2}}\left( 20 \right) \right)\left( {{\sin }^{2}}\left( 30 \right) \right)\left( {{\cos }^{2}}\left( 40 \right) \right)$
Using $\sin {{30}^{\circ }}=\dfrac{1}{2}$,we have
$\dfrac{1}{4}\left( {{\sin }^{2}}\left( 10 \right) \right)\left( {{\cos }^{2}}\left( 20 \right) \right)\left( {{\cos }^{2}}\left( 40 \right) \right)$
Now, we can club ${{\cos }^{2}}\left( {{20}^{\circ }} \right)$ and ${{\cos }^{2}}\left( {{40}^{\circ }} \right)$ together and use above mentioned expansions to simplify it further
$\dfrac{1}{4}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( \cos \left( 20 \right)\cos \left( 40 \right) \right)}^{2}}$
Multiplying and dividing by 2,
$\begin{align}
& =\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( 2\cos \left( 20 \right)\cos \left( 40 \right) \right)}^{2}} \\
& =\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( \cos \left( 40+20 \right)+\cos \left( 40-20 \right) \right)}^{2}} \\
& =\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right){{\left( \cos \left( 60 \right)+\cos \left( 20 \right) \right)}^{2}} \\
\end{align}$
Expanding the above expression using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$=\dfrac{1}{16}\left( {{\sin }^{2}}\left( 10 \right) \right)\left( {{\cos }^{2}}\left( 60 \right)+{{\cos }^{2}}\left( 20 \right)+2\cos \left( 60 \right)\cos \left( 20 \right) \right)$
Using the value of $\cos {{60}^{\circ }}=\dfrac{1}{2}$ and including ${{\sin }^{2}}\left( 10 \right)$ inside the brackets, we get
$=\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+{{\left( \sin \left( 10 \right)\cos \left( 20 \right) \right)}^{2}}+\sin \left( 10 \right)\sin \left( 10 \right)\cos \left( 20 \right) \right)$
Now, consider $\sin \left( 10 \right)\cos \left( 20 \right)$
Multiplying and dividing by 2, we get
$\dfrac{2\sin \left( 10 \right)\cos \left( 20 \right)}{2}$
Using the above expansions, this expression will be simplified as
$\begin{align}
& =\dfrac{\sin \left( 10+20 \right)+\sin \left( 10-20 \right)}{2} \\
& =\dfrac{\sin \left( 30 \right)-\sin \left( 10 \right)}{2} \\
\end{align}$
Using the value of $\sin {{30}^{\circ }}=\dfrac{1}{2}$, we have
$=\dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2}$
Now using this value in the above formulated expression, we have
$\begin{align}
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+{{\left( \sin \left( 10 \right)\cos \left( 20 \right) \right)}^{2}}+\sin \left( 10 \right)\sin \left( 10 \right)\cos \left( 20 \right) \right) \\
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+{{\left( \dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2} \right)}^{2}}+\sin \left( 10 \right)\left( \dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2} \right) \right) \\
\end{align}$
Expanding the expression, we get
$\begin{align}
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+\left( \dfrac{1}{16}+\dfrac{{{\sin }^{2}}\left( 10 \right)}{4}-\dfrac{\sin \left( 10 \right)}{4} \right)+\sin \left( 10 \right)\left( \dfrac{1}{4}-\dfrac{\sin \left( 10 \right)}{2} \right) \right) \\
& =\dfrac{1}{16}\left( \dfrac{{{\sin }^{2}}\left( 10 \right)}{4}+\dfrac{1}{16}+\dfrac{{{\sin }^{2}}\left( 10 \right)}{4}-\dfrac{\sin \left( 10 \right)}{4}+\dfrac{\sin \left( 10 \right)}{4}-\dfrac{{{\sin }^{2}}\left( 10 \right)}{2} \right) \\
& =\dfrac{1}{16}\times \dfrac{1}{16} \\
& =\dfrac{1}{256} \\
\end{align}$
Hence, the final value of $\sin \left( 10 \right)\cos \left( 20 \right)\sin \left( 30 \right)\cos \left( 40 \right)\sin \left( 50 \right)\cos \left( 60 \right)\sin \left( 70 \right)\cos \left( 80 \right)$ is $\dfrac{1}{256}$ .
Note: While solving such large expressions, always keep note of the constants that are being multiplied or divided in the expression because any change in them will change the whole answer. Also, apply each trigonometric identity carefully keeping in mind the positive and negative signs.
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