Question

If \[\sin \alpha =\dfrac{1}{2}\] then prove that \[3\cos \alpha -4{{\cos }^{3}}\alpha =0\].

Answer 1

Hint: In this question, we need to find the value of the cosine function from the given sine function using the trigonometric identity, Then consider the left hand side of the given expression in the question and substitute the value of the cosine function. Now, on solving it further we get the result.

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Complete step-by-step answer:

Now, from the given condition in the question we have

\[\Rightarrow \sin \alpha =\dfrac{1}{2}\]

As we already know that from the trigonometric identities the relation between sine and

cosine functions can be expressed as

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Now, by using the above identity we can say that

\[\Rightarrow {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\]

Now, this can also be written as

\[\Rightarrow {{\cos }^{2}}\alpha =1-{{\sin }^{2}}\alpha \]

Now, by substituting the value of sine function we get,

\[\Rightarrow {{\cos }^{2}}\alpha =1-{{\left( \dfrac{1}{2} \right)}^{2}}\]

Now, this can also be written as

\[\Rightarrow {{\cos }^{2}}\alpha =1-\dfrac{1}{4}\]

Now, on further simplification we get,

\[\Rightarrow {{\cos }^{2}}\alpha =\dfrac{3}{4}\]

Let us now apply the square root on both the sides

\[\therefore \cos \alpha =\dfrac{\sqrt{3}}{2}\]

Now, from the given expression in the question we have

\[\Rightarrow 3\cos \alpha -4{{\cos }^{3}}\alpha =0\]

Let us now consider the left hand side of the given expression

\[\Rightarrow 3\cos \alpha -4{{\cos }^{3}}\alpha \]

Now, on substituting the value of cosine function in the above expression we get,

\[\Rightarrow 3\left( \dfrac{\sqrt{3}}{2} \right)-4{{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}\]

Now, this can also be written further as

\[\Rightarrow \dfrac{3\sqrt{3}}{2}-4\left( \dfrac{3\sqrt{3}}{8} \right)\]

Now, on cancelling out the common terms we get,

\[\Rightarrow \dfrac{3\sqrt{3}}{2}-\dfrac{3\sqrt{3}}{2}\]

Now, on further simplification we get,

\[\Rightarrow 0\]

Hence, \[3\cos \alpha -4{{\cos }^{3}}\alpha =0\]



Note:
Instead of finding the cosine value from the sine value using trigonometric identity we can also find it by the trigonometric ratios formula. We need to find the hypotenuse, adjacent and opposite sides to get the value and then further substitute it in the given expression. Both the methods give the same result.

It is important to note that while finding the values we need to be careful about the signs and the calculations because neglecting any of the terms or considering incorrect values changes the result completely.