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

Answer: (C)

$0$

Answer 2

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.)