Question

The value of $${\cos ^2}{30^0} - {\sin ^2}{30^0}$$ is
$$\eqalign{
A. & \cos {60^0} \cr
B. & \sin {60^0} \cr
C. & 0 \cr
D. & 1 \cr} $$

Answer 1

Hint: We know the values of trigonometry term from trigonometry table. We need to put those values in given problems.

Complete step-by-step answer:
$$\eqalign{
  & {\cos ^2}{30^0} - {\sin ^2}{30^0} \cr
  & = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2} \cr
  & = \dfrac{3}{4} - \dfrac{1}{4} \cr
  & = \dfrac{1}{2} \cr
  & = \cos {60^0} \cr} $$
Hence, the answer is $$\cos {60^0}$$
Additional Information:
Trigonometry is a branch of mathematics that deals with triangles. Trigonometry is also known as the study of relationships between lengths and angles of triangles. It is used in various field:-
1.It is used in oceanography in calculating the height of tides in oceans.
2.The sine and cosine functions are fundamental to the theory of periodic functions, those that describe the sound and light waves.

Note: For solving these types of questions remember all trigonometric formulas and identities. This is a simple problem, just put the correct value to get an answer.