If tan θ = -4/3, then sin θ is (A) -4/5 but not 4/5 (B) -4/5 or 4/5 (C) 4/5 but not -4/5 (D) None of

Q: Wait, why can't we just use the identity $\sin^2 \theta + \cos^2 \theta = 1$?

You absolutely can! Using $\sec^2 \theta = 1 + \tan^2 \theta$ is a fantastic alternative. If you use $\tan^2 \theta = (-\frac{4}{3})^2 = \frac{16}{9}$, you get $\sec^2 \theta = 1 + \frac{16}{9} = \frac{25}{9}$. This means $\sec \theta = \pm \frac{5}{3}$, so $\cos \theta = \pm \frac{3}{5}$. Since $\sin \theta = \tan \theta \cdot \cos \theta$, you get $\sin \theta = (-\frac{4}{3}) \cdot (\pm \frac{3}{5}) = \mp \frac{4}{5}$. The identity works perfectly, but you must remember that the $\pm$ sign is not just a formality—it represents the two distinct quadrants where the tangent is negative.

Q: Why is the hypotenuse $r$ always positive?

Think of $r$ as the distance from the origin $(0,0)$ to the point $(x,y)$ on the terminal side of the angle. By the distance formula, $r = \sqrt{x^2 + y^2}$. Since $x^2$ and $y^2$ are always non-negative, their sum is non-negative, and the principal square root is defined as positive. Distance is a physical magnitude, and in this coordinate geometry context, it acts as the radius of a circle, which is always positive.

Q: How do I know if I should reject one of the answers?

In this specific problem, you shouldn't! The question asks for the value of $\sin \theta$ given only $\tan \theta = -\frac{4}{3}$. It does not restrict $\theta$ to a specific interval (like $0 < \theta < \pi$). Without that restriction, $\theta$ can be in any quadrant where $\tan \theta$ is negative. Therefore, both values are mathematically valid and must be included.

Question

If $	an 	heta = -4/3$, then $\sin 	heta$ is

Options:

(A) $-4/5$ but not $4/5$

(B) $-4/5$ or $4/5$

(C) $4/5$ but not $-4/5$

(D) None of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, fellow explorers of mathematics! Today, we are going to peel back the layers of a seemingly simple trigonometric problem. You might look at $ an heta = -\frac{4}{3}$ and think it is just a quick calculation, but there is a hidden geometric reality that separates the casual student from the master. Imagine you are standing at the origin of a Cartesian plane with an angle $ heta$ rotating around this center. The tangent function is defined as the ratio of the vertical displacement $y$ to the horizontal displacement $x$: $$ an heta = \frac{y}{x} $$ ### The Compass of the Coordinate Plane We are given $ an heta = -\frac{4}{3}$. The negative sign is our first clue, indicating that $x$ and $y$ must have opposite signs. This immediately restricts our angle $ heta$ to two specific regions: the second quadrant (where $x < 0, y > 0$) and the fourth quadrant (where $x > 0, y < 0$). This follows the **ASTC rule** (All, Sin, Tan, Cos). Since our tangent is negative, we are excluded from the first and third quadrants, leaving us only with the second and fourth. ### The Geometry of the Second Quadrant Let us step into the second quadrant, where $x$ is negative and $y$ is positive. We model the ratio $\frac{y}{x} = -\frac{4}{3}$ by setting $y = 4$ and $x = -3$. We calculate the hypotenuse $r$ using the Pythagorean theorem: $$ r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$ Now, we define $\sin heta$ as the ratio of the vertical component to the hypotenuse: $$ \sin heta = \frac{y}{r} = \frac{4}{5} $$ This positive result is consistent with the fact that sine is positive in the second quadrant. ### The Geometry of the Fourth Quadrant We must also consider the fourth quadrant, where $x$ is positive and $y$ is negative. To satisfy $\frac{y}{x} = -\frac{4}{3}$, we set $x = 3$ and $y = -4$. The hypotenuse remains constant: $$ r = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5 $$ Calculating the sine ratio for this quadrant yields: $$ \sin heta = \frac{y}{r} = \frac{-4}{5} = -\frac{4}{5} $$ This negative result is exactly what we expect in the fourth quadrant, demonstrating the elegant symmetry of the trigonometric functions. ### The Synthesis: Why Both Answers Matter We have found two distinct possibilities for $\sin heta$: $\frac{4}{5}$ and $-\frac{4}{5}$. In the context of advanced mathematics, we must be comprehensive as the problem does not restrict $ heta$ to a specific range. Therefore, the final answer is: $$ \sin heta = \pm \frac{4}{5} $$ Mathematics is not just about finding "the" answer; it is about understanding the landscape of all possible solutions. By navigating the quadrants and respecting the signs, you have arrived at the complete truth.

Animated Solution for Mathematics - Trigonometry: If $tan θ = - 4/3$ , then $sin θ$ is

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Visualized Solution

Analyzing $tan θ = - \frac{4}{3}$

Analyzing the Sign of $tan θ$

The ASTC Rule

Identifying the Quadrants for $θ$

Case 1: The 2nd Quadrant Setup

Calculating the Hypotenuse $r$ (Case 1)

Evaluating the Hypotenuse $r$ (Case 1)

Finding $sin θ$ in the 2nd Quadrant

Case 2: The 4th Quadrant Setup

Calculating the Hypotenuse $r$ (Case 2)

Finding $sin θ$ in the 4th Quadrant

The Final Conclusion for $sin θ$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Compass of the Coordinate Plane

The Geometry of the Second Quadrant

The Geometry of the Fourth Quadrant

The Synthesis

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If tanθ=−4/3, then sinθ is

Select Answer:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ=−34​

Analyzing the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ

The ASTC Rule

Identifying the Quadrants for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Case 1: The 2nd Quadrant Setup

Calculating the Hypotenuse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r (Case 1)

Evaluating the Hypotenuse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r (Case 1)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ in the 2nd Quadrant

Case 2: The 4th Quadrant Setup

Calculating the Hypotenuse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r (Case 2)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ in the 4th Quadrant

The Final Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ