Solve the given trigonometric expression $\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}=$
\[\begin{align}
& A.0 \\
& B.1 \\
& C.-1 \\
& D.2 \\
\end{align}\]
Answer
640.5k+ views
Hint: In this question, we are given an expression in terms of sine and cosine and we have to find its value without actually determining the values of individual function. For this we will use various trigonometric properties to reach our answer. Trigonometric properties that we will use are:
\[\begin{align}
& \left( i \right)1+\cos 2\theta =2{{\cos }^{2}}\theta \\
& \left( ii \right)\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right) \\
& \left( iii \right)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
& \left( iv \right)\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \\
& \left( v \right)\cos \left( -\theta \right)=\cos \theta \\
\end{align}\]
Complete step-by-step solution
Here we are given expression as
\[\Rightarrow \cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}=\]
For making some angles equal, as we can see half of angle of $\cos {{56}^{\circ }}$ will be $\cos {{28}^{\circ }}$. So let us first add and subtract 1 from expression, we get:
\[\Rightarrow 1+\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Now, we can apply $1+\cos 2\theta =2{{\cos }^{2}}\theta $ on $1+\cos {{56}^{\circ }}$ we get:
\[\Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Now let us apply the formula of difference of cosine functions given by $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$ on $\cos {{58}^{\circ }}-\cos {{66}^{\circ }}$ we get:
\[\Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\sin \left( \dfrac{58+66}{2} \right)\sin \left( \dfrac{66-58}{2} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Simplifying the angles of the sine function, we get:
\[\Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\sin {{62}^{\circ }}\sin {{4}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
As we can see, one of the cosine angle is ${{28}^{\circ }}$ and one of the sine angle is ${{62}^{\circ }}$ so let us change sine to cosine using $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ so that, we can get $\cos {{28}^{\circ }}$ we get:
\[\begin{align}
& \Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\cos \left( {{90}^{\circ }}-{{62}^{\circ }} \right)\sin {{4}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\cos {{28}^{\circ }}\sin {{4}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
Taking $\cos {{28}^{\circ }}$ common from first two terms, we get:
\[\Rightarrow 2\cos {{28}^{\circ }}\left( \cos {{28}^{\circ }}\sin {{4}^{\circ }} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Since, there is no formula for $\cos C+\sin D$ so let us change $\sin {{4}^{\circ }}$ to cosine function using transformation $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ we get:
\[\begin{align}
& \Rightarrow 2\cos {{28}^{\circ }}\left( \cos {{28}^{\circ }}+\cos \left( {{90}^{\circ }}-{{4}^{\circ }} \right) \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 2\cos {{28}^{\circ }}\left( \cos {{28}^{\circ }}+\cos {{86}^{\circ }} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
Applying $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ on $\cos {{28}^{\circ }}+\cos {{86}^{\circ }}$ we get:
\[\begin{align}
& \Rightarrow 2\cos {{28}^{\circ }}\left( 2\cos \left( \dfrac{{{28}^{\circ }}+{{86}^{\circ }}}{2} \right) \cos \left( \dfrac{{{28}^{\circ }}-{{86}^{\circ }}}{2} \right) \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 2\cos {{28}^{\circ }}\left( 2\cos {{57}^{\circ }} \cos \left( -{{29}^{\circ }} \right) \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
As we know, $\cos \left( -\theta \right)=\cos \theta $ so applying it on $\cos \left( -{{29}^{\circ }} \right)$ we get:
\[\begin{align}
& \Rightarrow 2\cos {{28}^{\circ }}\left( 2\cos {{57}^{\circ }}\cos {{29}^{\circ }} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
As we can see, in first two terms only difference is $\cos {{57}^{\circ }}\text{ and }\sin {{33}^{\circ }}$. Also, we can see that ${{90}^{\circ }}-{{57}^{\circ }}={{33}^{\circ }}$ so let us apply $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ on $\sin {{33}^{\circ }}$ we get:
\[\begin{align}
& \Rightarrow 4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos \left( {{90}^{\circ }}-{{33}^{\circ }} \right)-1 \\
& \Rightarrow 4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-1 \\
\end{align}\]
Cancelling first and second term we get: -1
Hence, the expression is reduced to -1.
Hence, value of $\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}$ is -1.
So option C is the correct answer.
Note: Students should take care of signs while applying the difference of cosine property. Students can make mistakes of positive, negative signs in trigonometric identities such as $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta ,\cos \left( -\theta \right)=\cos \theta $. While calculating angle after applying sum or difference of cosine formula, make sure that you divide the sum or the difference of angles by 2.
\[\begin{align}
& \left( i \right)1+\cos 2\theta =2{{\cos }^{2}}\theta \\
& \left( ii \right)\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right) \\
& \left( iii \right)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
& \left( iv \right)\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \\
& \left( v \right)\cos \left( -\theta \right)=\cos \theta \\
\end{align}\]
Complete step-by-step solution
Here we are given expression as
\[\Rightarrow \cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}=\]
For making some angles equal, as we can see half of angle of $\cos {{56}^{\circ }}$ will be $\cos {{28}^{\circ }}$. So let us first add and subtract 1 from expression, we get:
\[\Rightarrow 1+\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Now, we can apply $1+\cos 2\theta =2{{\cos }^{2}}\theta $ on $1+\cos {{56}^{\circ }}$ we get:
\[\Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Now let us apply the formula of difference of cosine functions given by $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$ on $\cos {{58}^{\circ }}-\cos {{66}^{\circ }}$ we get:
\[\Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\sin \left( \dfrac{58+66}{2} \right)\sin \left( \dfrac{66-58}{2} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Simplifying the angles of the sine function, we get:
\[\Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\sin {{62}^{\circ }}\sin {{4}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
As we can see, one of the cosine angle is ${{28}^{\circ }}$ and one of the sine angle is ${{62}^{\circ }}$ so let us change sine to cosine using $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ so that, we can get $\cos {{28}^{\circ }}$ we get:
\[\begin{align}
& \Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\cos \left( {{90}^{\circ }}-{{62}^{\circ }} \right)\sin {{4}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 2{{\cos }^{2}}{{28}^{\circ }}+2\cos {{28}^{\circ }}\sin {{4}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
Taking $\cos {{28}^{\circ }}$ common from first two terms, we get:
\[\Rightarrow 2\cos {{28}^{\circ }}\left( \cos {{28}^{\circ }}\sin {{4}^{\circ }} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1\]
Since, there is no formula for $\cos C+\sin D$ so let us change $\sin {{4}^{\circ }}$ to cosine function using transformation $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ we get:
\[\begin{align}
& \Rightarrow 2\cos {{28}^{\circ }}\left( \cos {{28}^{\circ }}+\cos \left( {{90}^{\circ }}-{{4}^{\circ }} \right) \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 2\cos {{28}^{\circ }}\left( \cos {{28}^{\circ }}+\cos {{86}^{\circ }} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
Applying $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ on $\cos {{28}^{\circ }}+\cos {{86}^{\circ }}$ we get:
\[\begin{align}
& \Rightarrow 2\cos {{28}^{\circ }}\left( 2\cos \left( \dfrac{{{28}^{\circ }}+{{86}^{\circ }}}{2} \right) \cos \left( \dfrac{{{28}^{\circ }}-{{86}^{\circ }}}{2} \right) \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 2\cos {{28}^{\circ }}\left( 2\cos {{57}^{\circ }} \cos \left( -{{29}^{\circ }} \right) \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
As we know, $\cos \left( -\theta \right)=\cos \theta $ so applying it on $\cos \left( -{{29}^{\circ }} \right)$ we get:
\[\begin{align}
& \Rightarrow 2\cos {{28}^{\circ }}\left( 2\cos {{57}^{\circ }}\cos {{29}^{\circ }} \right)-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
& \Rightarrow 4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}-1 \\
\end{align}\]
As we can see, in first two terms only difference is $\cos {{57}^{\circ }}\text{ and }\sin {{33}^{\circ }}$. Also, we can see that ${{90}^{\circ }}-{{57}^{\circ }}={{33}^{\circ }}$ so let us apply $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ on $\sin {{33}^{\circ }}$ we get:
\[\begin{align}
& \Rightarrow 4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos \left( {{90}^{\circ }}-{{33}^{\circ }} \right)-1 \\
& \Rightarrow 4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\cos {{57}^{\circ }}-1 \\
\end{align}\]
Cancelling first and second term we get: -1
Hence, the expression is reduced to -1.
Hence, value of $\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}$ is -1.
So option C is the correct answer.
Note: Students should take care of signs while applying the difference of cosine property. Students can make mistakes of positive, negative signs in trigonometric identities such as $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta ,\cos \left( -\theta \right)=\cos \theta $. While calculating angle after applying sum or difference of cosine formula, make sure that you divide the sum or the difference of angles by 2.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

