Question

The value of the expression $\dfrac{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }}-\sqrt{2}\sin {{27}^{\circ }} \right)\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }}+\sqrt{2}\sin {{27}^{\circ }} \right)}{2\sin {{54}^{\circ }}}$ is less than

& A.\cos {{36}^{\circ }} \\\ & B.\cos 67{{\dfrac{1}{2}}^{\circ }} \\\ & C.\cos {{18}^{\circ }} \\\ & D.\cos {{15}^{\circ }} \\\ \end{aligned}$$

Answer 1

At first, use the identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Then, expand using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and then, further use the formula ${{\sin }^{2}}\theta =2\sin \theta \cos \theta$. Then, further use ${{\cos }^{2}}\theta =1-2{{\sin }^{2}}\theta$ to simplify it and then, take the help of identity $\sin \left( {{90}^{\circ }}-\theta \right)$. After this, finally use the identity $\sin A+\sin B=2\sin \dfrac{\left( A+B \right)}{2}\cos \dfrac{\left( A-B \right)}{2}$.
Hence, use the fact that, if $\cos {{\theta }_{1}}<\cos {{\theta }_{2}}\text{ then }{{\theta }_{1}}>{{\theta }_{2}}$ or vice versa.

Complete step-by-step answer:
In the question, we are given an expression,
$\dfrac{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }}-\sqrt{2}\sin {{27}^{\circ }} \right)\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }}+\sqrt{2}\sin {{27}^{\circ }} \right)}{2\sin {{54}^{\circ }}}$
and we have to say that, which one of the following has lesser value than the given expression in the question.
Now, we will here proceed by using the following identity, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$and use it as $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ where we will take $\sin {{36}^{\circ }}+\cos {{36}^{\circ }}$ as a and $\sqrt{2}\sin {{27}^{\circ }}$ as b.
So, according to this, we can write the given expression as,

& \dfrac{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }}-\sqrt{2}\sin {{27}^{\circ }} \right)\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }}+\sqrt{2}\sin {{27}^{\circ }} \right)}{2\sin {{54}^{\circ }}} \\\ & \Rightarrow \dfrac{{{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }} \right)}^{2}}-2{{\sin }^{2}}{{27}^{\circ }}}{2\sin {{54}^{\circ }}} \\\ \end{aligned}$$ Now, we will expand $${{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }} \right)}^{2}}$$ using the identity $${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$$. So, we get, $${{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }} \right)}^{2}}$$ as $${{\sin }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{36}^{\circ }}+2\sin {{36}^{\circ }}\cos {{36}^{\circ }}$$ as we know that, $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$$ so, we can express the expansion as, $$1+2\sin {{36}^{\circ }}\cos {{36}^{\circ }}$$ we know that, $${{\sin }{2}}\theta =2\sin \theta \cos \theta $$ so, instead of $\theta $ if we take 36, then, we can write the given expansion as $$1+\sin \left( 2\times {{36}^{\circ }} \right)\Rightarrow 1+\sin {{72}^{\circ }}.$$ Now, let’s come back to the given simplified expression which was, $$\dfrac{{{\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }} \right)}^{2}}-2{{\sin }^{2}}{{27}^{\circ }}}{2\sin {{54}^{\circ }}}$$ Substituting the obtained value of $$\left( \sin {{36}^{\circ }}+\cos {{36}^{\circ }} \right)=1+\sin {{72}^{\circ }}$$ we get, $$\dfrac{1+\sin {{72}^{\circ }}-2{{\sin }^{2}}{{27}^{\circ }}}{2\sin {{54}^{\circ }}}$$ Rearranging the terms in the above expression, we get $$\dfrac{\sin {{72}^{\circ }}+1-2{{\sin }^{2}}{{27}^{\circ }}}{2\sin {{54}^{\circ }}}$$ Now, here we can use the identity that, $$1-2{{\sin }^{2}}\theta $$ Here,$$\theta $$ is $${{27}^{\circ }}$$. So, we can write it as, $$\begin{aligned} & \dfrac{\sin {{72}^{\circ }}+\cos \left( 2\times {{27}^{\circ }} \right)}{2\sin {{54}^{\circ }}} \\\ & \Rightarrow \dfrac{\sin {{72}^{\circ }}+\cos {{54}^{\circ }}}{2\sin {{54}^{\circ }}} \\\ \end{aligned}$$ Here, we will use an identity $$\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta .$$ so, instead of $$\theta $$ will use$${{54}^{\circ }}$$. So, $\sin \left( {{90}^{\circ }}-{{54}^{\circ }} \right)=\cos {{54}^{\circ }}\Rightarrow \sin {{36}^{\circ }}=\cos {{54}^{\circ }}.$ Hence, the expression becomes, $$\dfrac{\sin {{72}^{\circ }}+\sin {{36}^{\circ }}}{2\sin {{54}^{\circ }}}$$ Now, here we will use the identity that, $$\sin A+\sin B=2\sin \dfrac{\left( A+B \right)}{2}\cos \dfrac{\left( A-B \right)}{2}$$ Here, A is $${{72}^{\circ }}$$ and B is $${{36}^{\circ }}$$ so, we get the expression as, $$\begin{aligned} & \dfrac{2\sin \left( \dfrac{{{72}^{\circ }}+{{36}^{\circ }}}{2} \right)\cos \left( \dfrac{{{72}^{\circ }}-{{36}^{\circ }}}{2} \right)}{2\sin {{54}^{\circ }}} \\\ & \Rightarrow \dfrac{2\sin {{54}^{\circ }}\cos {{18}^{\circ }}}{2\sin {{54}^{\circ }}} \\\ \end{aligned}$$ Thus, the value of expression is $${{18}^{\circ }}$$ here, we will use the fact that if, $$\begin{aligned} & \cos {{\theta }_{1}}<\cos {{\theta }_{2}}\text{ then }{{\theta }_{1}}>{{\theta }_{2}} \\\ & \Rightarrow \cos {{\theta }_{1}}>\cos {{\theta }_{2}}\text{ then }{{\theta }_{1}}<{{\theta }_{2}} \\\ \end{aligned}$$ So, according to this only $$\cos {{15}^{\circ }}$$ is more than $$\cos {{18}^{\circ }}.$$ Hence, the correct option is D. **Note:** Students will generally make the mistake by finding out the expression value as $$\cos {{18}^{\circ }}$$ and then without reading the whole question they will mark option C instead of option D. Thus, one should be careful about that.