What is the difference between maximum and minimum value of expression y=1+2sinx+3cos^2x

To understand the difference between the maximum and minimum values of the expression $y=1+2\sin x+3{\cos }^{2}x$, we can break it down into manageable steps. This will involve using some properties of trigonometric functions and basic algebra.

Breaking Down the Expression

The expression can be rewritten using the identity ${\cos }^{2}x=1-{\sin }^{2}x$. This allows us to express everything in terms of sine, making it easier to analyze:

• Starting with $y=1+2\sin x+3{\cos }^{2}x$
• Substituting ${\cos }^{2}x$ gives us: $y=1+2\sin x+3(1-{\sin }^{2}x)$
• This simplifies to: $y=1+2\sin x+3-3{\sin }^{2}x$
• Thus, we have: $y=4+2\sin x-3{\sin }^{2}x$

Identifying the Structure

Now, we observe that $y$ is a quadratic function in terms of $\sin x$. We can let $u=\sin x$ (where $u$ ranges from -1 to 1), transforming our expression into:

$y=-3u^2+2u+4$

Finding Maximum and Minimum Values

This is a downward-opening parabola (since the coefficient of $u^2$ is negative), meaning it will have a maximum value at its vertex. The vertex of a quadratic function $a{x}^2+bx+c$ is found using the formula $u=-\frac{b}{2a}$.

In our case:

• $a=-3$
• $b=2$

Calculating the vertex:

$u=-\frac{2}{2\cdot -3}=\frac{1}{3}$

Calculating Maximum Value

Now, substitute $u=\frac{1}{3}$ back into the expression for $y$:

$y=-3{\left(\frac{1}{3}\right)}^2+2\left(\frac{1}{3}\right)+4$

Calculating this gives:

• $y=-3\cdot \frac{1}{9}+\frac{2}{3}+4$
• $y=-\frac{1}{3}+\frac{2}{3}+4=\frac{1}{3}+4=\frac{13}{3}$

Calculating Minimum Value

Now, we must check the endpoints of the interval $u=\sin x$. The possible maximum and minimum values occur at $u=-1$ and $u=1$. Let's calculate those:

For $u=-1$:

$y=-3(-1{)}^2+2(-1)+4=-3-2+4=-1$

For $u=1$:

$y=-3(1{)}^2+2(1)+4=-3+2+4=3$

Final Values and Their Difference

Now we have:

• Maximum value: $\frac{13}{3}\approx 4.33$
• Minimum value: $-1$

The difference between the maximum and minimum values is:

Difference = Maximum - Minimum = $\frac{13}{3}-(-1)=\frac{13}{3}+\frac{3}{3}=\frac{16}{3}$

So, the difference between the maximum and minimum values of the expression $y=1+2\sin x+3{\cos }^{2}x$ is $\frac{16}{3}$, or approximately 5.33.