Question

For any real number \(x\), the least value of \(4cosx-3sinx+5\) is ?

• A. \(10\)
• B. \(2\)
• C. \(0\)
• D. \(8\)
• E. \(4\)

Answer 1

The Correct Option is C

Given that:

\(4cosx-3sinx+5\)

for which we need to find the least value for any real number \(x\).

So we can proceed by using Pythagoras identity for cosine i.e. 

\(cos²(x) + sin²(x) = 1\)

Now, let's find the angle whose cosine is 4/5 and sine is 3/5. Let θ be that angle:

\(cos(θ) = 4/5\) , \(sin(θ) = 3/5\)

Using the Pythagorean identity:

\(cos²(θ) + sin²(θ) = 1\)

\( (4/5)² + (3/5)² = 1 \)

\(16/25 + 9/25 = 1\)

\( 25/25 = 1\)

Now, we have θ such that \(cos(θ) = 4/5\) and \(sin(θ) = 3/5\). 

So, \(4cos(x) - 3sin(x)\) can be rewritten as:

\(4cos(x) - 3sin(x) = 4cos(θ) - 3sin(θ)\)

Using the angle sum identity for cosine:

\(4cos(θ) - 3sin(θ) = 5cos(θ + α)\)

where α is the angle such that :

\(cos(α) = 3/5\) and \(sin(α) = 4/5\)

By again using Pythagorean identity we can write:

\(cos²(α) + sin²(α) = 1\)

\((3/5)² + (4/5)² = 1 \)

\(9/25 + 16/25 = 1 \)

\(25/25 = 1\)

So, we have α such that \(cos(α) = 3/5\) and \(sin(α) = 4/5\).

Now,\( 4cos(x) - 3sin(x) = 5cos(θ + α)\)

The least value of \(5cos(θ + α) \) is \(-5\), at \(cos(θ + α) = -1\).

Therefore, the least value of \(4cos(x) - 3sin(x) + 5\) is:

\(4cos(x) - 3sin(x) + 5 = 5cos(θ + α) + 5 = -5 + 5 = 0\)

So, the least value of the expression is \(0\) (_Ans.)