Question

If \[A=\sin {{45}^{\circ }}\sin {{12}^{\circ }};B=\cos {{45}^{\circ }}\cos {{12}^{\circ }};C=\cos {{66}^{\circ }}+\sin {{84}^{\circ }},\] then descending order of these values is
A. C, A, B
B. C, B, A
C. A, C, B
D. A, B, C

Answer 1

Hint: Value of \[\sin {{45}^{\circ }}\] and \[\cos {{45}^{\circ }}\] are \[\dfrac{1}{\sqrt{2}}\] of each. \[\sin \theta \] Is increasing in domain \[\left[ 0,\dfrac{\pi }{2} \right]\] and \[\cos \theta \] is decreasing in the same domain. Value of \[\sin \theta =0,\sin \dfrac{\pi }{2}=1,\cos 0=1,\cos \dfrac{\pi }{2}=0\] and \[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] . If we multiply any number with a number between \[\left( 0,1 \right)\] , the number will become smaller, but if the number is multiplied with a number that is greater than 0, the new number becomes larger than the previous number. Use these concepts to solve the problem.

Complete step-by-step answer:
We are given values A, B, C in the problem are

 \[A=\sin {{45}^{\circ }}\sin {{12}^{\circ }}\] \[\to \] (1)
 \[B=\cos {{45}^{\circ }}\cos {{12}^{\circ }}\] \[\to \] (2)
 \[C=\cos {{66}^{\circ }}+\sin {{84}^{\circ }}\] \[\to \] (3)

We know value of \[\sin {{45}^{\circ }}\] and \[\cos {{45}^{\circ }}\] are \[\dfrac{1}{\sqrt{2}}\] of each.
Hence, we can put \[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\] in equation (1). We get
 \[A=\dfrac{1}{\sqrt{2}}\sin {{12}^{\circ }}\] \[\to \] (4)

Similarly, we can get value of B as

 \[B=\cos {{45}^{\circ }}\cos {{12}^{\circ }}\]

 \[B=\dfrac{1}{\sqrt{2}}\cos {{12}^{\circ }}\]

 \[B=\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}\] \[\to \] (5)

Now, as we know \[\sin \theta \] is an increasing function from 0 to \[{{90}^{\circ }}\] and value of \[\sin {{0}^{\circ }}=0\] ,
 \[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\] And \[\sin {{90}^{\circ }}=1\] .
One other hand \[\cos \theta \] is decreasing function and values of \[\cos {{0}^{\circ }}=1\] , \[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\] and \[\cos {{90}^{\circ }}=0\] .
It means \[\sin \theta >\cos \theta \] for the range \[{{45}^{\circ }}\] to \[{{90}^{\circ }}\] and \[\cos \theta >\sin \theta \] for the range \[{{0}^{\circ }}\] to \[{{45}^{\circ }}\] .
So, we get,

 \[\cos {{12}^{\circ }}>\sin {{12}^{\circ }}\]

Multiply by \[\dfrac{1}{\sqrt{2}}\] to both sides of the above equation. We get

 \[\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}>\dfrac{\sin {{12}^{\circ }}}{\sqrt{2}}\]

Here \[B>A\] \[\to \] (6)
Now, we have

 \[C=\cos {{66}^{\circ }}+\sin {{84}^{\circ }}\]

We know

 \[\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \]

Put \[\theta ={{6}^{\circ }}\] , we get

 \[\sin \left( {{90}^{\circ }}-6 \right)=\cos {{6}^{\circ }}\]

 \[\sin {{84}^{\circ }}=\cos {{6}^{\circ }}\]

Hence, we can rewrite expression ‘C’ as

 \[C=\cos {{66}^{\circ }}+\cos {{6}^{\circ }}\]

We know the trigonometric identity of \[\cos x+\cos y\] , can be given as

 \[\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\]

So, we can simplify ‘C’ as

 \[C=2\cos \left( \dfrac{66+6}{2} \right)\cos \left( \dfrac{66-6}{2} \right)\]

 \[C=2\cos {{36}^{\circ }}\cos {{30}^{\circ }}\]
We know \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
So, we get

 \[C=\dfrac{2\sqrt{3}}{2}\cos {{36}^{\circ }}\]

 \[C=\sqrt{3}\cos {{36}^{\circ }}\] \[\to \] (7)

So, as we know cosine is a decreasing function for \[\left[ 0,{{90}^{\circ }} \right]\] .
So, we get
 \[\cos {{12}^{\circ }}>\cos {{36}^{\circ }}\]
Now, if we multiply \[\cos {{12}^{\circ }}\] by \[\dfrac{1}{\sqrt{2}}\] (smaller than 1), the term \[\dfrac{1}{\sqrt{2}}\cos {{12}^{\circ }}\] becomes less and as \[\dfrac{1}{\sqrt{2}}=\cos {{45}^{\circ }}\] , it means the term \[\cos {{45}^{\circ }}\cos {{12}^{\circ }}\] will be less than \[\cos {{45}^{\circ }}\] as \[\cos {{12}^{\circ }}\] will also belong to \[\left( 0,1 \right)\] . It means the term \[\dfrac{1}{\sqrt{2}}\cos {{12}^{\circ }}\] will be less than \[\cos {{36}^{\circ }}\] as well, if \[\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}\] is less than \[\cos {{45}^{\circ }}\] as cosine is a decreasing function. So, we get

 \[\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}<\cos {{36}^{\circ }}\]

Or, \[\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}<\sqrt{3}\cos {{36}^{\circ }}\]
We multiplied by \[\sqrt{3}\] , as \[\sqrt{3}\] is greater than 1, so, \[\sqrt{3}\cos {{36}^{\circ }}\] will be higher than \[\cos {{36}^{\circ }}\] , hence no change in inequality.
Hence, we get

 \[\dfrac{\sin {{12}^{\circ }}}{\sqrt{2}}<\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}<\sqrt{3}\cos {{36}^{\circ }}\]
 \[\begin{align}
  & A < B < C \\
 & or \\
 & C > B > A \\
\end{align}\]

Hence, the decreasing order of A, B, C is C, B, A. So, option (b) is correct.

So, the correct answer is “Option (b)”.

Note: Relating \[\cos {{45}^{\circ }}\cos {{12}^{\circ }}\] and \[\sqrt{3}\cos {{36}^{\circ }}\] is the key point of the question.
It uses the concept that if any number gets multiplied by a number less than 1, then the result becomes less than the number and if any number is multiplied with a number greater than 1, then the number will become larger than the previous number.
One may try to calculate exact values of the given expressions, but it is a really difficult and complex approach. So, don’t try to calculate exact values of them, just relate them.