Question

What is the value of $cos\left( 36{}^\circ -A \right)cos\left( 36+A \right)+\cos \left( 54{}^\circ +A \right)\cos \left( 54{}^\circ -A \right)$ .

Answer 1

Hint:Use the property that trigonometric ratios are periodic functions and the knowledge of complementary angles related to trigonometric ratios. Then finally, use the formula of cos(A-B) to reach an answer.

Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.
Next, let us see the graph of cosx.
Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ$ . So, we can mathematically show it as:
$\sin (2{{\pi }^{c}}+x)=\sin (360{}^\circ +x)=\sin x$
$\cos (2{{\pi }^{c}}+x)=cos(360{}^\circ +x)=\cos x$
Also, some other results to remember are:
$\sin \left( \dfrac{{{\pi }^{c}}}{2}-x \right)=\sin (90{}^\circ -x)=\cos x$
$\cos \left( \dfrac{{{\pi }^{c}}}{2}-x \right)=\cos (90{}^\circ -x)=\sin x$
We will now solve the expression given in the question.
$\therefore cos\left( 36{}^\circ -A \right)cos\left( 36+A \right)+\cos \left( 54{}^\circ +A \right)\cos \left( 54{}^\circ -A \right)$
$=cos\left( 36{}^\circ -A \right)cos\left( 36+A \right)+\cos \left( 90{}^\circ -\left( 36{}^\circ -A \right) \right)\cos \left( 90{}^\circ -\left( 36{}^\circ +A \right) \right)$
We know $\cos \left( \dfrac{{{\pi }^{c}}}{2}-x \right)=\cos (90{}^\circ -x)=\sin x$ , so, our expression becomes:
$cos\left( 36{}^\circ -A \right)cos\left( 36+A \right)+\cos \left( 90{}^\circ -\left( 36{}^\circ -A \right) \right)\cos \left( 90{}^\circ -\left( 36{}^\circ +A \right) \right)$
$=cos\left( 36{}^\circ -A \right)cos\left( 36+A \right)+sin\left( 36{}^\circ -A \right)\sin \left( 36{}^\circ +A \right)$
We know cos(A-B) = cosAcosB + sinAsinB, applying this to our expression, we get
$\cos (36{}^\circ +A-(36{}^\circ -A))$
$=\cos (2A)$
Therefore, the answer to the above question is cos2A.

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, remember the property of complementary angles of trigonometric ratios.