If $\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = \operatorname{tg} 45^{\circ}$, What Is The Value Of $B = \operatorname{tg}(\frac{3 \theta}{2}) \cdot \cos (2 \theta)$?

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In the realm of trigonometry, where angles and their relationships intertwine, we often encounter intriguing problems that demand a blend of trigonometric identities, algebraic manipulation, and insightful reasoning. This article delves into one such problem, where we are given the condition tg⁡θ⋅tg⁡2θ=tg⁡45∘\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = \operatorname{tg} 45^{\circ} and tasked with finding the value of the expression B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}\left(\frac{3 \theta}{2}\right) \cdot \cos (2 \theta). Let's embark on this mathematical journey, unraveling the solution step by step.

Unveiling the Trigonometric Identity

At the heart of this problem lies a fundamental trigonometric identity: tg⁡45∘=1\operatorname{tg} 45^{\circ} = 1. This seemingly simple equation holds the key to unlocking the value of B. By substituting this identity into the given condition, we arrive at a crucial starting point:

tg⁡θ⋅tg⁡2θ=1\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = 1

This equation establishes a relationship between the tangent of angle θ and the tangent of twice that angle. To further explore this connection, we can express tg⁡2θ\operatorname{tg} 2 \theta in terms of tg⁡θ\operatorname{tg} \theta using the double-angle formula:

tg⁡2θ=2tg⁡θ1−tg⁡2θ\operatorname{tg} 2 \theta = \frac{2 \operatorname{tg} \theta}{1 - \operatorname{tg}^2 \theta}

Substituting this expression into our equation, we obtain:

tg⁡θ⋅2tg⁡θ1−tg⁡2θ=1\operatorname{tg} \theta \cdot \frac{2 \operatorname{tg} \theta}{1 - \operatorname{tg}^2 \theta} = 1

Navigating the Algebraic Maze

Now, we're faced with an algebraic equation involving tg⁡θ\operatorname{tg} \theta. To simplify this equation, let's introduce a substitution: let x=tg⁡θx = \operatorname{tg} \theta. Our equation then transforms into:

x⋅2x1−x2=1x \cdot \frac{2x}{1 - x^2} = 1

Multiplying both sides by (1−x2)(1 - x^2), we get:

2x2=1−x22x^2 = 1 - x^2

Rearranging the terms, we arrive at a quadratic equation:

3x2=13x^2 = 1

Solving for x2x^2, we find:

x2=13x^2 = \frac{1}{3}

Taking the square root of both sides, we obtain two possible values for x:

x=±13x = \pm \frac{1}{\sqrt{3}}

Since x=tg⁡θx = \operatorname{tg} \theta, this implies that tg⁡θ=±13\operatorname{tg} \theta = \pm \frac{1}{\sqrt{3}}.

Deciphering the Angle θ

The values of tg⁡θ\operatorname{tg} \theta we've obtained correspond to specific angles. Recall that tg⁡30∘=13\operatorname{tg} 30^{\circ} = \frac{1}{\sqrt{3}} and tg⁡150∘=−13\operatorname{tg} 150^{\circ} = -\frac{1}{\sqrt{3}}. Therefore, the possible values for θ are:

θ=30∘\theta = 30^{\circ} or θ=150∘\theta = 150^{\circ}

However, we must be cautious and check if both solutions are valid. If θ=150∘\theta = 150^{\circ}, then 2θ=300∘2\theta = 300^{\circ}, and tg⁡2θ=tg⁡300∘=−13\operatorname{tg} 2\theta = \operatorname{tg} 300^{\circ} = -\frac{1}{\sqrt{3}}. In this case, tg⁡θ⋅tg⁡2θ=(−13)⋅(−13)=13\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = \left(-\frac{1}{\sqrt{3}}\right) \cdot \left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{3}, which does not satisfy the initial condition tg⁡θ⋅tg⁡2θ=1\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = 1. Therefore, θ=150∘\theta = 150^{\circ} is not a valid solution.

Thus, we conclude that θ=30∘\theta = 30^{\circ} is the only valid solution.

Calculating the Value of B

Now that we've determined the value of θ, we can proceed to calculate the value of B. Recall that:

B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}\left(\frac{3 \theta}{2}\right) \cdot \cos (2 \theta)

Substituting θ=30∘\theta = 30^{\circ}, we get:

B=tg⁡(3⋅30∘2)⋅cos⁡(2⋅30∘)B = \operatorname{tg}\left(\frac{3 \cdot 30^{\circ}}{2}\right) \cdot \cos (2 \cdot 30^{\circ})

Simplifying the expression, we have:

B=tg⁡45∘⋅cos⁡60∘B = \operatorname{tg} 45^{\circ} \cdot \cos 60^{\circ}

We know that tg⁡45∘=1\operatorname{tg} 45^{\circ} = 1 and cos⁡60∘=12\cos 60^{\circ} = \frac{1}{2}. Therefore:

B=1⋅12=12B = 1 \cdot \frac{1}{2} = \frac{1}{2}

However, upon reviewing the options provided (A. 2, B. 3, C. 6, D. Discussion), we realize that 12\frac{1}{2} is not among the choices. This indicates a potential error in the problem statement or the given options. Let's re-examine our steps to ensure accuracy.

After careful review, we identify a mistake in the final calculation. While the individual values of tg⁡45∘\operatorname{tg} 45^{\circ} and cos⁡60∘\cos 60^{\circ} are correct, their product was incorrectly calculated. The correct calculation is:

B=1⋅12=12B = 1 \cdot \frac{1}{2} = \frac{1}{2}

This discrepancy highlights the importance of meticulousness in mathematical problem-solving. Even a small error can lead to an incorrect answer.

Given the options provided, none of them match the correct value of B=12B = \frac{1}{2}. Therefore, the most appropriate response would be to indicate that none of the given options are correct.

Conclusion

Through a combination of trigonometric identities, algebraic manipulation, and careful analysis, we've navigated the intricacies of this problem. We determined that the value of θ\theta that satisfies the given condition is 30∘30^{\circ}, and subsequently calculated the value of the expression B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}\left(\frac{3 \theta}{2}\right) \cdot \cos (2 \theta) to be 12\frac{1}{2}. Although this value does not match any of the provided options, the process underscores the power of trigonometric principles and the importance of precision in mathematical reasoning. This exploration exemplifies how a seemingly complex problem can be unraveled through a systematic approach, highlighting the beauty and elegance inherent in the world of mathematics.

Let's delve into this intriguing trigonometric question. We're given the equation tg⁡θ⋅tg⁡2θ=tg⁡45∘\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = \operatorname{tg} 45^{\circ} and tasked with finding the value of B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}(\frac{3 \theta}{2}) \cdot \cos (2 \theta). To solve this, we will leverage trigonometric identities and algebraic manipulation. This journey will not only reveal the answer but also provide a deeper appreciation for the elegance of trigonometry.

The first step in tackling this problem involves simplifying the given condition. Recall that the tangent of 45 degrees is equal to 1, that is, tg⁡45∘=1\operatorname{tg} 45^{\circ} = 1. Therefore, the initial equation becomes:

tg⁡θ⋅tg⁡2θ=1\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = 1

This equation sets the stage for our investigation, highlighting the relationship between the tangent of an angle and the tangent of its double. To further dissect this relationship, we employ the double-angle formula for the tangent function:

tg⁡2θ=2tg⁡θ1−tg⁡2θ\operatorname{tg} 2 \theta = \frac{2 \operatorname{tg} \theta}{1 - \operatorname{tg}^2 \theta}

This formula allows us to express the tangent of 2θ2\theta in terms of tg⁡θ\operatorname{tg} \theta, enabling us to consolidate the equation into a single variable. Now, substitute the double-angle formula into the initial equation:

tg⁡θ⋅2tg⁡θ1−tg⁡2θ=1\operatorname{tg} \theta \cdot \frac{2 \operatorname{tg} \theta}{1 - \operatorname{tg}^2 \theta} = 1

Transforming the Trigonometric Expression

Now, let's simplify this equation to solve for θ\theta. This is where the elegance of algebraic manipulation shines. To make the equation more manageable, we introduce a substitution. Let x=tg⁡θx = \operatorname{tg} \theta. The equation then transforms into:

x⋅2x1−x2=1x \cdot \frac{2x}{1 - x^2} = 1

Multiplying both sides by (1−x2)(1 - x^2) to eliminate the fraction gives us:

2x2=1−x22x^2 = 1 - x^2

Combining like terms, we get a quadratic equation:

3x2=13x^2 = 1

Solving for x2x^2, we find:

x2=13x^2 = \frac{1}{3}

Taking the square root of both sides, we obtain two potential solutions for x:

x=±13x = \pm \frac{1}{\sqrt{3}}

Since x=tg⁡θx = \operatorname{tg} \theta, these solutions imply that:

tg⁡θ=13\operatorname{tg} \theta = \frac{1}{\sqrt{3}} or tg⁡θ=−13\operatorname{tg} \theta = -\frac{1}{\sqrt{3}}

These values correspond to angles where the tangent function takes on these specific values. Remembering our unit circle and special angles, we recall that tg⁡30∘=13\operatorname{tg} 30^{\circ} = \frac{1}{\sqrt{3}} and tg⁡150∘=−13\operatorname{tg} 150^{\circ} = -\frac{1}{\sqrt{3}}. Therefore, potential values for θ\theta are 30∘30^{\circ} and 150∘150^{\circ}.

However, we must exercise caution and verify that both potential solutions are valid within the original equation. Let's check θ=150∘\theta = 150^{\circ} first. If θ=150∘\theta = 150^{\circ}, then 2θ=300∘2\theta = 300^{\circ}, and tg⁡2θ=tg⁡300∘=−13\operatorname{tg} 2\theta = \operatorname{tg} 300^{\circ} = -\frac{1}{\sqrt{3}}. Plugging these values into the original equation, we get:

tg⁡150∘⋅tg⁡300∘=(−13)⋅(−13)=13\operatorname{tg} 150^{\circ} \cdot \operatorname{tg} 300^{\circ} = \left(-\frac{1}{\sqrt{3}}\right) \cdot \left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{3}

Since 13≠1\frac{1}{3} \neq 1, θ=150∘\theta = 150^{\circ} is not a valid solution. This step is crucial because it highlights the importance of checking for extraneous solutions that may arise during the algebraic manipulation process.

Therefore, the only valid solution for θ\theta is 30∘30^{\circ}.

Final Calculation of the Trigonometric Value

With the value of θ\theta securely in hand, we can now calculate the value of B. Recall that:

B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}\left(\frac{3 \theta}{2}\right) \cdot \cos (2 \theta)

Substituting θ=30∘\theta = 30^{\circ}, we have:

B=tg⁡(3⋅30∘2)⋅cos⁡(2⋅30∘)B = \operatorname{tg}\left(\frac{3 \cdot 30^{\circ}}{2}\right) \cdot \cos (2 \cdot 30^{\circ})

Simplifying the expression, we get:

B=tg⁡45∘⋅cos⁡60∘B = \operatorname{tg} 45^{\circ} \cdot \cos 60^{\circ}

Now, we know the values of these trigonometric functions for these special angles. The tangent of 45 degrees is 1, and the cosine of 60 degrees is 12\frac{1}{2}. Therefore:

B=1⋅12=12B = 1 \cdot \frac{1}{2} = \frac{1}{2}

Thus, the value of the expression B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}(\frac{3 \theta}{2}) \cdot \cos (2 \theta) is 12\frac{1}{2}.

In conclusion, by applying trigonometric identities, algebraic manipulation, and careful validation of solutions, we've successfully determined the value of B. This journey underscores the beauty and interconnectedness of mathematical concepts and the importance of rigorous problem-solving techniques. The answer, B=12B = \frac{1}{2}, although not matching the provided options, serves as a testament to the accuracy of our method and the importance of critical evaluation in mathematics.

Trigonometric equations often appear daunting, but with the right approach and a solid understanding of trigonometric identities, they can be solved systematically. In this article, we'll tackle the equation tg⁡θ⋅tg⁡2θ=tg⁡45∘\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = \operatorname{tg} 45^{\circ} and use it to find the value of B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}(\frac{3 \theta}{2}) \cdot \cos (2 \theta). This will be a journey through key trigonometric principles, algebraic techniques, and careful verification.

Laying the Foundation: Trigonometric Identities and Simplification

The first step in solving any trigonometric problem is to simplify the given information. Here, we have tg⁡θ⋅tg⁡2θ=tg⁡45∘\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = \operatorname{tg} 45^{\circ}. A crucial trigonometric identity to recall is that tg⁡45∘=1\operatorname{tg} 45^{\circ} = 1. This simplifies our equation to:

tg⁡θ⋅tg⁡2θ=1\operatorname{tg} \theta \cdot \operatorname{tg} 2 \theta = 1

This equation establishes a relationship between the tangent of θ{\theta} and the tangent of twice the angle (2θ{\theta}). To explore this relationship further, we'll utilize the double-angle formula for the tangent function:

tg⁡2θ=2tg⁡θ1−tg⁡2θ\operatorname{tg} 2 \theta = \frac{2 \operatorname{tg} \theta}{1 - \operatorname{tg}^2 \theta}

This formula is a cornerstone of trigonometry, allowing us to express the tangent of a doubled angle in terms of the tangent of the original angle. Substituting this into our simplified equation, we get:

tg⁡θ⋅2tg⁡θ1−tg⁡2θ=1\operatorname{tg} \theta \cdot \frac{2 \operatorname{tg} \theta}{1 - \operatorname{tg}^2 \theta} = 1

Now, we have an equation solely in terms of tg⁡θ\operatorname{tg} \theta, which paves the way for algebraic manipulation.

Algebraic Maneuvers: Solving for Tangent

To solve for tg⁡θ\operatorname{tg} \theta, we introduce a substitution. Let x=tg⁡θx = \operatorname{tg} \theta. This transforms our equation into:

x⋅2x1−x2=1x \cdot \frac{2x}{1 - x^2} = 1

Multiplying both sides by (1−x2)(1 - x^2) eliminates the fraction, resulting in:

2x2=1−x22x^2 = 1 - x^2

Rearranging the terms, we obtain a quadratic equation:

3x2=13x^2 = 1

Dividing both sides by 3, we get:

x2=13x^2 = \frac{1}{3}

Taking the square root of both sides yields two potential solutions:

x=±13x = \pm \frac{1}{\sqrt{3}}

Since x=tg⁡θx = \operatorname{tg} \theta, we have two possible values for the tangent of θ{\theta}:

tg⁡θ=13\operatorname{tg} \theta = \frac{1}{\sqrt{3}} or tg⁡θ=−13\operatorname{tg} \theta = -\frac{1}{\sqrt{3}}

These values correspond to specific angles. We know that tg⁡30∘=13\operatorname{tg} 30^{\circ} = \frac{1}{\sqrt{3}} and tg⁡150∘=−13\operatorname{tg} 150^{\circ} = -\frac{1}{\sqrt{3}}. Therefore, possible values for θ{\theta} are 30° and 150°.

Verification and the Importance of Context

However, it's crucial to verify these solutions. Plugging θ=150∘\theta = 150^{\circ} back into the original equation, we find:

tg⁡150∘⋅tg⁡(2⋅150∘)=tg⁡150∘⋅tg⁡300∘=(−13)⋅(−13)=13≠1\operatorname{tg} 150^{\circ} \cdot \operatorname{tg} (2 \cdot 150^{\circ}) = \operatorname{tg} 150^{\circ} \cdot \operatorname{tg} 300^{\circ} = \left(-\frac{1}{\sqrt{3}}\right) \cdot \left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{3} \neq 1

Therefore, θ=150∘\theta = 150^{\circ} is an extraneous solution and must be discarded. This emphasizes the importance of checking solutions in trigonometric equations to ensure they satisfy the original condition. The only valid solution is θ=30∘\theta = 30^{\circ}.

The Final Calculation: Finding the Value of B

Now that we have the correct value for θ{\theta}, we can find the value of B:

B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}\left(\frac{3 \theta}{2}\right) \cdot \cos (2 \theta)

Substituting θ{\theta} = 30°, we get:

B=tg⁡(3⋅30∘2)⋅cos⁡(2⋅30∘)B = \operatorname{tg}\left(\frac{3 \cdot 30^{\circ}}{2}\right) \cdot \cos (2 \cdot 30^{\circ})

Simplifying, we have:

B=tg⁡45∘⋅cos⁡60∘B = \operatorname{tg} 45^{\circ} \cdot \cos 60^{\circ}

We know that tg⁡45∘=1\operatorname{tg} 45^{\circ} = 1 and cos⁡60∘=12\cos 60^{\circ} = \frac{1}{2}. Thus:

B=1⋅12=12B = 1 \cdot \frac{1}{2} = \frac{1}{2}

Therefore, the value of B=tg⁡(3θ2)⋅cos⁡(2θ)B = \operatorname{tg}(\frac{3 \theta}{2}) \cdot \cos (2 \theta) is 12\frac{1}{2}. This result, while definitive, highlights a potential issue in the original problem statement if the answer choices provided did not include this value. This situation underscores the importance of rigorous problem-solving and the need to critically evaluate the context of the problem.

In summary, we successfully solved the trigonometric equation by applying key identities, performing algebraic manipulations, verifying solutions, and calculating the final value of B. This journey showcases the power of a systematic approach and the beauty inherent in trigonometric relationships.