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

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

$\dfrac{\sin {{12}^{\circ }}}{\sqrt{2}}<\dfrac{\cos {{12}^{\circ }}}{\sqrt{2}}<\sqrt{3}\cos {{36}^{\circ }}$

& A < B < C \\\ & or \\\ & C > B > A \\\ \end{aligned}$$ Hence, the decreasing order of A, B, C is C, B, A. So, option (b) is correct. **So, the correct answer is “Option (b)”.** **Note:** Relating $$\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.