Problem 34

Question

$$ \text { If the angle } \alpha \text { is in the third quadrant and } \tan \alpha=2 \text { , then find the value of } \sin \alpha \text { . } $$

Step-by-Step Solution

Verified

Answer

$\sin\alpha = -\frac{2}{\sqrt{5}}$

1Step 1: Find the value of cos(α) using tan(α)

We know that tan(α) = sin(α)/cos(α). We are given the value of tan(α), which is 2. So, $ 2 = \frac{\sin\alpha}{\cos\alpha} $ Now, we need to find the value of cos(α). To do this, we can rewrite the equation: $\cos\alpha = \frac{\sin\alpha}{2}$

2Step 2: Find sin²(α) using the trigonometric identity

We know that sin²(α) + cos²(α) = 1. We can now substitute the expression for cos(α) that we found in step 1 into this equation: $\sin^2\alpha + \left(\frac{\sin\alpha}{2}\right)^2 = 1$

3Step 3: Solve the quadratic equation for sin(α)

Simplify the equation from step 2: $\sin^2\alpha + \frac{\sin^2\alpha}{4} = 1$ Combine the terms with sin²(α): $\frac{5}{4}\sin^2\alpha = 1$ Now, solve for sin²(α): $\sin^2\alpha = \frac{4}{5}$

4Step 4: Find the value of sin(α)

Since α is in the third quadrant, sin(α) is negative. Therefore, we will take the negative square root of the value we found for sin²(α) in step 3: $\sin\alpha = -\sqrt{\frac{4}{5}} = -\frac{2}{\sqrt{5}}$ Thus, the value of sin(α) is: $\sin\alpha = -\frac{2}{\sqrt{5}}$

Key Concepts

Trigonometric functionsQuadratic equationsTrigonometric quadrants

Trigonometric functions

Trigonometric functions are essential tools in mathematics that relate the angles and sides of triangles. The three primary trigonometric functions are sine ($\sin$), cosine ($\cos$), and tangent ($\tan$). Each function has a specific meaning when analyzing right triangles or angles in the unit circle.

**Sine ($\sin$)**: This function represents the ratio of the opposite side to the hypotenuse in a right triangle.
**Cosine ($\cos$)**: This indicates the ratio of the adjacent side to the hypotenuse.
**Tangent ($\tan$)**: This describes the ratio of the opposite side to the adjacent side, or equivalently, $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$.

These functions are not only pivotal in geometry but are also used widely in calculus and physics for modeling periodic phenomena such as waves.

Quadratic equations

Quadratic equations are polynomial equations of the second degree, usually in the form $ax^2 + bx + c = 0$. They can be solved using various methods like factoring, completing the square, or applying the quadratic formula. In our context, when dealing with trigonometric identities, translation of identities to a quadratic form helps in solving complex problems in simpler ways.
In the given exercise, we saw that by using the identity $\sin^2\alpha + \cos^2\alpha = 1$, we arrived at a quadratic equation format with $\sin^2\alpha$. Solving:

Expressing both sine and cosine in terms of one variable simplifies complex trigonometric problems.
Recognizing patterns and forms such as $\sin^2\alpha + \left(\frac{\sin\alpha}{2}\right)^2 = 1$, allows the transformation into a quadratic equation $\frac{5}{4} \sin^2\alpha = 1$, ensuring easy solving using known algebraic techniques.

Trigonometric quadrants

Trigonometric quadrants divide the coordinate plane into four parts, each one representing a specific range of angles:

First quadrant: All trigonometric functions are positive.
Second quadrant: Sine is positive, while cosine and tangent are negative.
Third quadrant: Both sine and cosine are negative, but tangent is positive.
Fourth quadrant: Cosine is positive, while sine and tangent are negative.

Understanding these quadrants is essential in correctly evaluating and interpreting trigonometric functions like sine, cosine, and tangent.
In the exercise for $\alpha$ located in the third quadrant, it is crucial to note that $\sin(\alpha)$ is negative, which affects the final solution. Knowing which functions are positive or negative in each quadrant will help verify if your final answer makes sense. Thus, when solving $sin(\alpha)$, recognizing that it must be negative due to its position in the third quadrant is a handy rule to remember.

Problem 33

Problem 35

Other exercises in this chapter

Problem 32

$$ (\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}=\tan ^{2} A+\cot ^{2} A+7 $$

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Problem 33

$$ (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\cos e c A}{\sec ^{2} A} $$

View solution

Problem 35

$$ \text { If } \theta \text { is an acute angle and } \tan \theta=\frac{1}{\sqrt{7}}, \text { then find the value of } \frac{\cos e c^{2} \theta-\sec ^{2} \the

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Problem 36

$$ \text { If } \tan \theta=\frac{p}{q}, \text { show that } \frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}=\frac{p^{2}-q^{2}}{p^{2}+q^{2}} \te

View solution