Solving the Trigonometric Equation $\sin 4A - \cos 4A = 1$

The problem asks us to find the value of $A/2$ in degrees, given the equation $\sin 4A - \cos 4A = 1$ and the constraint $0 < A \le 90^\circ$. We are also provided with options for the value of $A/2$.

The provided options for $A/2$ are 60, 30, 40, and 45 degrees. Let's consider what these values imply for $A$:

• If $A/2 = 60^\circ$, then $A = 120^\circ$. This is outside the given range $0 < A \le 90^\circ$.
• If $A/2 = 30^\circ$, then $A = 60^\circ$. This is within the range $0 < A \le 90^\circ$.
• If $A/2 = 40^\circ$, then $A = 80^\circ$. This is within the range $0 < A \le 90^\circ$.
• If $A/2 = 45^\circ$, then $A = 90^\circ$. This is within the range $0 < A \le 90^\circ$.

Checking the Given Equation $\sin 4A - \cos 4A = 1$

Let's test the values of $A$ derived from the options that are within the allowed range ($60^\circ, 80^\circ, 90^\circ$) in the original equation $\sin 4A - \cos 4A = 1$.

• If $A = 60^\circ$: $4A = 4 \times 60^\circ = 240^\circ$. $\sin 240^\circ - \cos 240^\circ = \left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}\right) = \frac{1 - \sqrt{3}}{2}$. This is not equal to 1. So $A/2 = 30^\circ$ is not the solution for the given equation.
• If $A = 80^\circ$: $4A = 4 \times 80^\circ = 320^\circ$. $\sin 320^\circ - \cos 320^\circ = \sin(360^\circ - 40^\circ) - \cos(360^\circ - 40^\circ) = -\sin 40^\circ - \cos 40^\circ$. Since $\sin 40^\circ$ and $\cos 40^\circ$ are both positive, this value is negative and cannot be equal to 1. So $A/2 = 40^\circ$ is not the solution for the given equation.
• If $A = 90^\circ$: $4A = 4 \times 90^\circ = 360^\circ$. $\sin 360^\circ - \cos 360^\circ = 0 - 1 = -1$. This is not equal to 1. So $A/2 = 45^\circ$ is not the solution for the given equation $\sin 4A - \cos 4A = 1$.

None of the potential values of $A$ derived from the options satisfy the given equation $\sin 4A - \cos 4A = 1$. The provided correct answer corresponds to $A/2 = 45^\circ$, which means $A = 90^\circ$. However, as shown above, $A = 90^\circ$ yields $\sin 360^\circ - \cos 360^\circ = -1$, not 1.

Solving a Related Trigonometric Equation

It appears there might be a discrepancy in the problem statement or the options. Let's consider if the intended equation might have been slightly different, specifically if the right-hand side was -1 instead of 1, as $A=90^\circ$ resulted in -1.

Consider the equation: $\sin 4A - \cos 4A = -1$.

To solve this, we can express the left side in the form $R \sin(4A - \alpha)$ or $R \cos(4A + \alpha)$. We calculate $R = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.

Divide the equation by $\sqrt{2}$:

$\frac{1}{\sqrt{2}} \sin 4A - \frac{1}{\sqrt{2}} \cos 4A = -\frac{1}{\sqrt{2}}$

Using $\frac{1}{\sqrt{2}} = \cos 45^\circ = \sin 45^\circ$, we can write:

$\sin 4A \cos 45^\circ - \cos 4A \sin 45^\circ = -\sin 45^\circ$

Using the trigonometric identity $\sin(X - Y) = \sin X \cos Y - \cos X \sin Y$, where $X=4A$ and $Y=45^\circ$:

$\sin(4A - 45^\circ) = -\sin 45^\circ$

Since $-\sin \theta = \sin(-\theta)$ and also $-\sin \theta = \sin(180^\circ + \theta)$, we can write:

$\sin(4A - 45^\circ) = \sin(-45^\circ)$ or $\sin(4A - 45^\circ) = \sin(180^\circ + 45^\circ) = \sin(225^\circ)$.

Using the general solution for $\sin x = \sin \alpha$, which is $x = n \cdot 180^\circ + (-1)^n \alpha$ (where $n$ is an integer):

Case 1: $\sin(4A - 45^\circ) = \sin(-45^\circ)$

$4A - 45^\circ = n \cdot 180^\circ + (-1)^n (-45^\circ)$

• If $n$ is even, $n=2k$: $4A - 45^\circ = 2k \cdot 180^\circ + (-1)^{2k} (-45^\circ)$ $4A - 45^\circ = 360k^\circ - 45^\circ$ $4A = 360k^\circ$ $A = 90k^\circ$ For $k=1$, $A = 90^\circ$. This is within the range $0 < A \le 90^\circ$. For $A=90^\circ$, $A/2 = 45^\circ$. For $k=0$, $A=0^\circ$, which is not within the range $0 < A$.
• If $n$ is odd, $n=2k+1$: $4A - 45^\circ = (2k+1) \cdot 180^\circ + (-1)^{2k+1} (-45^\circ)$ $4A - 45^\circ = 360k^\circ + 180^\circ + 45^\circ$ $4A - 45^\circ = 360k^\circ + 225^\circ$ $4A = 360k^\circ + 270^\circ$ $A = 90k^\circ + 67.5^\circ$ For $k=0$, $A = 67.5^\circ$. This is within the range $0 < A \le 90^\circ$. For $A=67.5^\circ$, $A/2 = 33.75^\circ$.

Case 2: $\sin(4A - 45^\circ) = \sin(225^\circ)$ (This leads to the same general solution structure as $\sin(-45^\circ)$ because $225^\circ = 180^\circ + 45^\circ$, and $\sin(180+\theta) = -\sin\theta$) $4A - 45^\circ = n \cdot 180^\circ + (-1)^n (225^\circ)$ This will yield the same set of solutions for A as in Case 1, but expressed differently through the index n. For instance, if $n=0$ in Case 2, $4A - 45^\circ = 225^\circ \implies 4A = 270^\circ \implies A = 67.5^\circ$. If $n=1$ in Case 2, $4A - 45^\circ = 180^\circ - 225^\circ = -45^\circ \implies 4A = 0^\circ \implies A = 0^\circ$ (not in range). If $n=2$ in Case 2, $4A - 45^\circ = 360^\circ + 225^\circ = 585^\circ \implies 4A = 630^\circ \implies A = 157.5^\circ$ (not in range). If $n=3$ in Case 2, $4A - 45^\circ = 540^\circ - 225^\circ = 315^\circ \implies 4A = 360^\circ \implies A = 90^\circ$.

The solutions for $A$ in the range $0 < A \le 90^\circ$ for the equation $\sin 4A - \cos 4A = -1$ are $A=67.5^\circ$ and $A=90^\circ$.

The corresponding values for $A/2$ are:

• If $A = 67.5^\circ$, then $A/2 = 33.75^\circ$.
• If $A = 90^\circ$, then $A/2 = 45^\circ$.

From the options provided for $A/2$ (60, 30, 40, 45), only 45 matches one of the solutions derived from the equation $\sin 4A - \cos 4A = -1$. Since the provided correct answer text is 45, it aligns with $A/2 = 45^\circ$, which is a valid solution if the equation was $\sin 4A - \cos 4A = -1$. Although the given equation is $\sin 4A - \cos 4A = 1$, we conclude that based on the provided options and correct answer, the intended problem likely leads to $A/2 = 45^\circ$. This occurs when $A=90^\circ$, which solves $\sin 4A - \cos 4A = -1$.

Conclusion for the Given Question

Based on the analysis, the value $A/2 = 45^\circ$ corresponds to $A=90^\circ$. While $A=90^\circ$ does not satisfy the equation $\sin 4A - \cos 4A = 1$, it satisfies $\sin 4A - \cos 4A = -1$. Given that $45^\circ$ is provided as the correct answer option for $A/2$, we select this value.

The value of $A/2$, in degrees, that aligns with the provided correct option is 45.

Revision Table: Key Concepts

Additional Information on Trigonometric Identities and Solving

Solving trigonometric equations often involves using identities to simplify the equation into a basic form like $\sin x = k$, $\cos x = k$, or $\tan x = k$, where $k$ is a constant. For expressions like $a \sin \theta + b \cos \theta$, they can be transformed into the form $R \sin(\theta + \alpha)$ or $R \cos(\theta - \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\alpha$ is an angle determined by $a$ and $b$ (e.g., $\cos \alpha = a/R, \sin \alpha = b/R$). This transformation helps in solving equations involving sums or differences of sine and cosine terms with the same angle.

When finding general solutions, the periodicity of trigonometric functions is crucial. Sine and cosine functions have a period of $360^\circ$ or $2\pi$ radians, while the tangent function has a period of $180^\circ$ or $\pi$ radians. The general solutions incorporate an integer $n$ to represent all possible angles.

It's important to always check if the derived solutions for the angle fall within any specified range given in the problem. If the problem asks for a value like $A/2$ or $2A$, remember to calculate that final value after finding $A$.