Question

If $ {{\cos }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\cos {{45}^{\circ }}\sin {{45}^{\circ }} $ then value of $ x $ is equal to \[\]
A.2\[\]
B. $ \dfrac{1}{2} $ \[\]
C. $ \dfrac{-1}{2} $ \[\]
D. $ -1 $ \[\]

Answer 1

Hint: We recall the definitions of sine and cosine trigonometric ratios in the right-angled triangle. We recall the values of sine and cosine for angles of measure $ {{30}^{\circ }},{{45}^{\circ }} $ . We put these values of sine and cosine in the given equation and then solve for $ x $ . \[\]

We know that in a right-angled triangle the side opposite to the right-angled triangle is called hypotenuse denoted as $ h $ , the vertical side is called perpendicular denoted as $ p $ and the horizontal side is called the base denoted as $ b $ .\[\]

Complete step by step answer:
We know from the trigonometric ratios in a right angled triangle the sine of any angle is given by the ratio of length of side opposite to the angle to the length of hypotenuse. In the figure the sine of the angle $ \theta $ is given by
\[\sin \theta =\dfrac{p}{h}\]
Similarly the cosine of an angle is the ratio of length of side adjacent to the angle (excluding hypotenuse) to the length of hypotenuse. So we have cosine of angle $ \theta $ as
\[\cos \theta =\dfrac{b}{h}\]
 We know from the basic sine and cosine values of angles of measures $ {{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }} $ as
\[\begin{align}
  & \sin {{30}^{\circ }}=\dfrac{1}{2},\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
 & \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{60}^{\circ }}=\dfrac{1}{2} \\
\end{align}\]
We are asked to find the values of $ x $ from the following equation,
\[\begin{align}
  & {{\cos }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\cos {{45}^{\circ }}\sin {{45}^{\circ }} \\
 & \Rightarrow {{\left( \cos {{45}^{\circ }} \right)}^{2}}-{{\left( \cos {{30}^{\circ }} \right)}^{2}}=x\cos {{45}^{\circ }}\sin {{45}^{\circ }} \\
\end{align}\]
We put the known values of sine and cosine in the above equation to have;
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=x\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right) \\
 & \Rightarrow \dfrac{1}{2}-\dfrac{3}{4}=x\dfrac{1}{\sqrt{2}\sqrt{2}} \\
 & \Rightarrow \dfrac{1-3}{4}=x\dfrac{1}{2} \\
 & \Rightarrow \dfrac{-1}{2}=x\dfrac{1}{2} \\
 & \Rightarrow x=-1 \\
\end{align}\]
So the option is D. \[\]

Note:
We should always remember that value of $ \sin {{45}^{\circ }},\cos {{45}^{\circ }} $ are equal. We should remember that if the degree symbols are not there we should always take the measure of the angle in radian for trigonometric values. We note that whatever may be the measure sine ad cosine will always lie between $ -1 $ and 1. We should remember the formula $ {{\cos }^{2}}A-{{\cos }^{2}}B=\sin \left( A+B \right)\sin \left( A-B \right) $ .