Solving the Trigonometric Equation and Finding sin2x

The problem asks us to consider a given trigonometric equation and use it to find the value of an expression involving parameters derived from $\sin 2x$. The given equation is:

$\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7$

We need to simplify this equation to relate it to $\sin 2x$. We can express all trigonometric functions in terms of $\sin x$ and $\cos x$:

$\sin x + \cos x + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\cos x} + \frac{1}{\sin x} = 7$

Combine the terms with common denominators:

$(\sin x + \cos x) + \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right) + \left(\frac{1}{\cos x} + \frac{1}{\sin x}\right) = 7$

$(\sin x + \cos x) + \left(\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\right) + \left(\frac{\sin x + \cos x}{\sin x \cos x}\right) = 7$

Using the identity $\sin^2 x + \cos^2 x = 1$ and the relation $\sin x \cos x = \frac{1}{2}\sin 2x$, we can simplify:

Let $s = \sin x + \cos x$ and $p = \sin x \cos x$. The equation becomes:

$s + \frac{1}{p} + \frac{s}{p} = 7$

$s + \frac{1+s}{p} = 7$

We also know that $s^2 = (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + 2p$. So, $p = \frac{s^2 - 1}{2}$.

Substitute $p$ into the simplified equation:

$s + \frac{1+s}{\frac{s^2 - 1}{2}} = 7$

$s + \frac{2(1+s)}{s^2 - 1} = 7$

Since $s^2 - 1 = (s-1)(s+1)$, we have:

$s + \frac{2(1+s)}{(s-1)(s+1)} = 7$

Assuming $s+1 \neq 0$ (which holds for the range of $s$), we can cancel the $(s+1)$ term:

$s + \frac{2}{s-1} = 7$

Multiply both sides by $(s-1)$ to clear the denominator:

$s(s-1) + 2 = 7(s-1)$

$s^2 - s + 2 = 7s - 7$

Rearrange into a quadratic equation in $s$:

$s^2 - s - 7s + 2 + 7 = 0$

$s^2 - 8s + 9 = 0$

Now, we solve this quadratic equation for $s = \sin x + \cos x$ using the quadratic formula $s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$s = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(9)}}{2(1)}$

$s = \frac{8 \pm \sqrt{64 - 36}}{2}$

$s = \frac{8 \pm \sqrt{28}}{2}$

$s = \frac{8 \pm \sqrt{4 \times 7}}{2}$

$s = \frac{8 \pm 2\sqrt{7}}{2}$

$s = 4 \pm \sqrt{7}$

The possible values for $s = \sin x + \cos x$ are $4 + \sqrt{7}$ and $4 - \sqrt{7}$. We know that the maximum value of $\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$ is $\sqrt{2} \approx 1.414$, and the minimum value is $-\sqrt{2} \approx -1.414$. The value of $4 + \sqrt{7} \approx 4 + 2.64 = 6.64$, which is outside the range $[-\sqrt{2}, \sqrt{2}]$. The value of $4 - \sqrt{7} \approx 4 - 2.64 = 1.36$, which is within the range $[-\sqrt{2}, \sqrt{2}]$.

Therefore, we must have $s = \sin x + \cos x = 4 - \sqrt{7}$.

Finding the Value of sin2x

We need to find $\sin 2x$. We know that $\sin 2x = 2 \sin x \cos x = (\sin x + \cos x)^2 - (\sin^2 x + \cos^2 x) = s^2 - 1$.

Using the value $s = 4 - \sqrt{7}$:

$\sin 2x = (4 - \sqrt{7})^2 - 1$

Expand the square:

$\sin 2x = (4^2 - 2(4)(\sqrt{7}) + (\sqrt{7})^2) - 1$

$\sin 2x = (16 - 8\sqrt{7} + 7) - 1$

$\sin 2x = (23 - 8\sqrt{7}) - 1$

$\sin 2x = 22 - 8\sqrt{7}$

Determining a, b, and c

The problem states that $\sin 2x = a - b\sqrt{c}$, where $a$ and $b$ are natural numbers and $c$ is a prime number.

Comparing our result $\sin 2x = 22 - 8\sqrt{7}$ with the form $a - b\sqrt{c}$:

$a = 22$

$b = 8$

$c = 7$

Let's check if these values meet the conditions:

• $a = 22$ is a natural number (1, 2, 3, ...). Yes.
• $b = 8$ is a natural number. Yes.
• $c = 7$ is a prime number (a natural number greater than 1 that has no positive divisors other than 1 and itself). Yes.

All conditions are satisfied.

Calculating the Final Expression

We need to find the value of $a - b + 2c$ using the values $a=22$, $b=8$, and $c=7$.

Value $= a - b + 2c = 22 - 8 + 2(7)$

Value $= 14 + 14$

Value $= 28$

Thus, the value of $a - b + 2c$ is 28.

Revision Table: Key Concepts

Additional Information: Simplifying Trigonometric Expressions

Simplifying complex trigonometric expressions is a common task in trigonometry. The main strategy is often to convert all terms into the base functions $\sin x$ and $\cos x$. This helps in finding common denominators and using fundamental identities like $\sin^2 x + \cos^2 x = 1$.

For expressions involving $\tan x, \cot x, \sec x, \csc x$, remember their definitions in terms of $\sin x$ and $\cos x$:

• $\tan x = \frac{\sin x}{\cos x}$
• $\cot x = \frac{\cos x}{\sin x}$
• $\sec x = \frac{1}{\cos x}$
• $\csc x = \frac{1}{\sin x}$

Also, sums like $\sin x + \cos x$ and products like $\sin x \cos x$ are often related to double angle formulas, such as $\sin 2x = 2 \sin x \cos x$. The square of the sum $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x$ is a very useful identity in these kinds of problems.

When solving equations involving sums and products of $\sin x$ and $\cos x$, substituting $s = \sin x + \cos x$ and $p = \sin x \cos x$ and using the relation $s^2 = 1 + 2p$ can simplify the equation significantly, often leading to polynomial equations that are easier to solve.