Question

Write 'True' or 'False' and justify your answer in the following: 

\[\cos \theta = \frac{a^2 + b^2}{2ab}\], where a and b are two distinct numbers such that ab > 0.

Answer 1

This statement is false.

Explanation:

It is given that, \[\sin\theta = x + \frac{1}{x}\]

\[\Rightarrow - 1 \leq x + \frac{1}{x} \leq 1\]

\[\Rightarrow x + \frac{1}{x} \leq 1\]

\[\Rightarrow x^2 + 1 \leq x\]

\[\Rightarrow x^2 + 1 - x \leq 0\]

\[\text{Take }x = 1, \]

\[ \Rightarrow 1 + 1 - 1 \leq 0\]

\[ \Rightarrow 1 \leq 0\]

Which is false, so x is not always a positive real number.

Answer 2

This statement is false.

Explanation:

Given: a ≠ b and ab > 0

(Because Arithmetic Mean (AM) of a list of non-negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

⇒ AM > GM

If a and b be such numbers, then

AM = `(a + b)/2` and Gm = `sqrt(ab)`

By assuming that cos θ = `(a^2 + b^2)/(2ab)` is true statement.

Similarly, AM and GM of a² and b² will be,

AM = `(a^2 + b^2)/2` and GM = `sqrt(a^2 * b^2)`

So, `(a^2 + b^2)/2 > sqrt(a^2 * b^2)`   ...(By AM and GM property as mentioned earlier in the answer)

⇒ `(a^2 + b^2)/2 > ab`

⇒ `(a^2 + b^2)/(2ab) > 1`

⇒ cos θ > 1  ...(By our assumption)

But this not possible since, –1 ≤ cos θ ≤ 1

Thus, our assumption is wrong and `cos theta ≠ (a^2 + b^2)/(2ab)`