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Question: Solve the given trigonometric expression \(\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\ci...

Solve the given trigonometric expression cos56+cos58cos664cos28cos29sin33=\cos {{56}^{\circ }}+\cos {{58}^{\circ }}-\cos {{66}^{\circ }}-4\cos {{28}^{\circ }}\cos {{29}^{\circ }}\sin {{33}^{\circ }}=

& A.0 \\\ & B.1 \\\ & C.-1 \\\ & D.2 \\\ \end{aligned}$$
Explanation

Solution

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:

& \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{aligned}$$ **Complete step-by-step solution** Here we are given expression as $$\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{aligned} & \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{aligned}$$ Taking $\cos {{28}^{\circ }}$ common from first two terms, we get: $$\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{aligned} & \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{aligned}$$ Applying $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ on $\cos {{28}^{\circ }}+\cos {{86}^{\circ }}$ we get: $$\begin{aligned} & \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{aligned}$$ As we know, $\cos \left( -\theta \right)=\cos \theta $ so applying it on $\cos \left( -{{29}^{\circ }} \right)$ we get: $$\begin{aligned} & \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{aligned}$$ As we can see, in first two terms only difference is $\cos {{57}^{\circ }}\text{ and }\sin {{33}^{\circ }}$. Also, we can see that ${{90}^{\circ }}-{{57}^{\circ }}={{33}^{\circ }}$ so let us apply $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ on $\sin {{33}^{\circ }}$ we get: $$\begin{aligned} & \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{aligned}$$ 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.