Question

Verify the following equalities $\cos 60^\circ = 1 - 2{\sin ^2}30^\circ = 2{\cos ^2}30^\circ - 1$

Answer 1

Hint: Here we are asked to check whether the given equalities are the same or not. As we can see that all the terms are given in trigonometric functions so first, we will try to find all the three values of the given expressions. For finding the values we will be using trigonometric ratios to simplify them and to find the values of them. Then we will compare all the values to check whether they all are the same or not.

Complete answer:
Since from given that $\cos 60^\circ = 1 - 2{\sin ^2}30^\circ = 2{\cos ^2}30^\circ - 1$ this means the equalities like $\cos 60^\circ = 1 - 2{\sin ^2}30^\circ $ , $1 - 2{\sin ^2}30^\circ = 2{\cos ^2}30^\circ - 1$ and $\cos 60^\circ = 2{\cos ^2}30^\circ - 1$ (all the three values are same)
Starting with the trigonometric table of sine and cosine to solve the given equalities.
Let us start with the sine table in the trigonometric functions with respect to the corresponding angles

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\sin \]	\[0\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	\[1\]

Similarly, cosine table in the trigonometric functions with respect to the corresponding angles

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\cos \]	\[1\]	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	\[0\]

Now take the first given equality value, which is $\cos 60^\circ $ and we can clearly see that $\cos 60^\circ = \dfrac{1}{2}$
Now take the second value from the given, that is $1 - 2{\sin ^2}30^\circ $
Since we know that $\sin 30^\circ = \dfrac{1}{2}$ and apply this value in the above we get \[1 - 2{\sin ^2}30^\circ = 1 - 2{(\dfrac{1}{2})^2}\]
Further solving we get \[1 - 2{\sin ^2}30^\circ = 1 - \dfrac{1}{2} \Rightarrow \dfrac{1}{2}\] and hence we have \[1 - 2{\sin ^2}30^\circ = \dfrac{1}{2}\]
Now take the final value $2{\cos ^2}30^\circ - 1$ and since we know that $\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$
Substitute the value in the above we get $2{\cos ^2}30^\circ - 1 = 2{(\dfrac{{\sqrt 3 }}{2})^2} - 1$
Further solving we get $2{\cos ^2}30^\circ - 1 = 2 \times \dfrac{3}{4} - 1 \Rightarrow \dfrac{1}{2}$ and hence we have $2{\cos ^2}30^\circ - 1 = \dfrac{1}{2}$
Therefore, all the three values are equals because $\cos 60^\circ = \dfrac{1}{2}$ , \[1 - 2{\sin ^2}30^\circ = \dfrac{1}{2}\] and $2{\cos ^2}30^\circ - 1 = \dfrac{1}{2}$
Hence $\cos 60^\circ = 1 - 2{\sin ^2}30^\circ = 2{\cos ^2}30^\circ - 1$

Note:
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $and $\tan = \dfrac{1}{{\cot }}$
And also, we know that the relation of the sine, cosine, and tangent is $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ and hence we get the tangent table as

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
$\tan $	\[0\]	\[\dfrac{1}{{\sqrt 3 }}\]	\[1\]	\[\sqrt 3 \]	undefined