Trigonometric Identities And Values Solving Problems Step-by-Step

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In the realm of mathematics, trigonometry stands as a pivotal branch that explores the relationships between angles and sides of triangles. Trigonometric identities and values form the bedrock of this field, enabling us to solve a myriad of problems in geometry, physics, engineering, and beyond. This comprehensive guide delves into the intricacies of trigonometric identities and values, equipping you with the knowledge and skills to master these fundamental concepts. This article aims to provide a detailed explanation of how to solve trigonometric problems, focusing on specific examples involving sine, cosine, and their relationships. We will explore how to determine trigonometric values using given information and identities, without relying on calculators. Understanding these principles is crucial for success in mathematics and related fields.

1.18.1 Problem Statement

The cornerstone of our exploration lies in the given equation:

sin⁡17∘cos⁡17∘=k\sin 17^\circ \cos 17^\circ = k

Our mission is to express the following trigonometric expressions in terms of k:

  1. sin⁡34∘\sin 34^\circ
  2. 2sin⁡217∘−12\sin^2 17^\circ - 1

1.18.2 Solution

1.18.2.1 Part 1: Expressing sin⁡34∘\sin 34^\circ in terms of k

To embark on this journey, we invoke the double-angle identity for sine, a fundamental cornerstone of trigonometric manipulations. This identity elegantly states that:

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta

In our specific scenario, we judiciously set θ\theta to 17∘17^\circ. This strategic substitution unveils a profound connection:

sin⁡(2×17∘)=2sin⁡17∘cos⁡17∘\sin (2 \times 17^\circ) = 2 \sin 17^\circ \cos 17^\circ

Simplifying the left-hand side, we arrive at:

sin⁡34∘=2sin⁡17∘cos⁡17∘\sin 34^\circ = 2 \sin 17^\circ \cos 17^\circ

Now, a pivotal moment arrives. We recognize that the right-hand side of this equation bears a striking resemblance to the expression given in the problem statement:

sin⁡17∘cos⁡17∘=k\sin 17^\circ \cos 17^\circ = k

With a flourish of algebraic substitution, we seamlessly replace 2sin⁡17∘cos⁡17∘2 \sin 17^\circ \cos 17^\circ with 2k2k, culminating in the elegant solution:

sin⁡34∘=2k\sin 34^\circ = 2k

This elegant result unveils the value of sin⁡34∘\sin 34^\circ expressed in terms of the given constant, k. Understanding and applying the double-angle identities is crucial for simplifying trigonometric expressions and solving problems efficiently. The ability to recognize and utilize these identities allows for a more streamlined approach to complex problems, making it easier to find solutions. In this case, recognizing the relationship between sin⁡34∘\sin 34^\circ and sin⁡17∘cos⁡17∘\sin 17^\circ \cos 17^\circ through the double-angle identity was the key to solving this part of the problem. The result, sin⁡34∘=2k\sin 34^\circ = 2k, not only provides the answer but also highlights the power and elegance of trigonometric identities.

1.18.2.2 Part 2: Expressing 2sin⁡217∘−12\sin^2 17^\circ - 1 in terms of k

For the second part of our problem, we aim to express 2sin⁡217∘−12\sin^2 17^\circ - 1 in terms of k. To achieve this, we'll leverage another fundamental trigonometric identity: the double-angle identity for cosine. There are three common forms of this identity, but the most relevant one for our purpose is:

cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2 \theta

This form directly relates cos⁡2θ\cos 2\theta to sin⁡2θ\sin^2 \theta, which aligns perfectly with our target expression. By rearranging this identity, we can isolate the term we're interested in:

2sin⁡2θ−1=−cos⁡2θ2\sin^2 \theta - 1 = -\cos 2\theta

Now, we set θ=17∘\theta = 17^\circ. Substituting this value into the rearranged identity gives us:

2sin⁡217∘−1=−cos⁡(2×17∘)2\sin^2 17^\circ - 1 = -\cos (2 \times 17^\circ)

Simplifying the angle on the right-hand side, we have:

2sin⁡217∘−1=−cos⁡34∘2\sin^2 17^\circ - 1 = -\cos 34^\circ

Our next step is to express cos⁡34∘\cos 34^\circ in terms of k. To do this, we'll use the Pythagorean identity, which relates sine and cosine:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

We can rearrange this identity to solve for cos⁡θ\cos \theta:

cos⁡θ=±1−sin⁡2θ\cos \theta = \pm \sqrt{1 - \sin^2 \theta}

Applying this to cos⁡34∘\cos 34^\circ, we get:

cos⁡34∘=±1−sin⁡234∘\cos 34^\circ = \pm \sqrt{1 - \sin^2 34^\circ}

From Part 1, we know that sin⁡34∘=2k\sin 34^\circ = 2k. Substituting this into the equation above gives us:

cos⁡34∘=±1−(2k)2=±1−4k2\cos 34^\circ = \pm \sqrt{1 - (2k)^2} = \pm \sqrt{1 - 4k^2}

Now we need to determine the sign of cos⁡34∘\cos 34^\circ. Since 34∘34^\circ is in the first quadrant, where cosine is positive, we take the positive square root:

cos⁡34∘=1−4k2\cos 34^\circ = \sqrt{1 - 4k^2}

Finally, we substitute this expression back into our equation for 2sin⁡217∘−12\sin^2 17^\circ - 1:

2sin⁡217∘−1=−cos⁡34∘=−1−4k22\sin^2 17^\circ - 1 = -\cos 34^\circ = -\sqrt{1 - 4k^2}

Thus, we have expressed 2sin⁡217∘−12\sin^2 17^\circ - 1 in terms of k. This solution demonstrates the importance of recognizing and applying trigonometric identities strategically. By using the double-angle identity for cosine and the Pythagorean identity, we were able to express the given expression in terms of k without needing a calculator. This approach underscores the elegance and power of trigonometric manipulations.

1.19.1 Problem Statement

Given that sin⁡A=23\sin A = \frac{2}{3} and cos⁡A<0\cos A < 0, our objective is to determine the values of other trigonometric functions of angle A without using a calculator. This type of problem emphasizes the understanding of trigonometric relationships and identities rather than relying on computational tools. The key is to utilize the given information and fundamental trigonometric identities to find the required values.

1.19.2 Solution

Step 1: Determine the Quadrant of Angle A

To begin, let's analyze the given information: sin⁡A=23\sin A = \frac{2}{3} and cos⁡A<0\cos A < 0. The sine function is positive in the first and second quadrants, while the cosine function is negative in the second and third quadrants. The only quadrant where both conditions are satisfied is the second quadrant. Therefore, angle A lies in the second quadrant. Understanding the quadrant in which the angle lies is crucial because it helps determine the signs of the trigonometric functions.

Step 2: Calculate cos⁡A\cos A using the Pythagorean Identity

We know that sin⁡A=23\sin A = \frac{2}{3}, and we need to find cos⁡A\cos A. We can use the Pythagorean identity, which states:

sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1

Substituting the given value of sin⁡A\sin A, we get:

(23)2+cos⁡2A=1\left(\frac{2}{3}\right)^2 + \cos^2 A = 1

Simplifying, we have:

49+cos⁡2A=1\frac{4}{9} + \cos^2 A = 1

Subtract 49\frac{4}{9} from both sides:

cos⁡2A=1−49=59\cos^2 A = 1 - \frac{4}{9} = \frac{5}{9}

Taking the square root of both sides, we get:

cos⁡A=±59=±53\cos A = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}

Since we know that cos⁡A<0\cos A < 0 in the second quadrant, we choose the negative value:

cos⁡A=−53\cos A = -\frac{\sqrt{5}}{3}

Step 3: Calculate tan⁡A\tan A

Now that we have sin⁡A\sin A and cos⁡A\cos A, we can find tan⁡A\tan A using the identity:

tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}

Substituting the values of sin⁡A\sin A and cos⁡A\cos A, we get:

tan⁡A=23−53=23×3−5=−25\tan A = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = \frac{2}{3} \times \frac{3}{-\sqrt{5}} = -\frac{2}{\sqrt{5}}

To rationalize the denominator, we multiply the numerator and denominator by 5\sqrt{5}:

tan⁡A=−25×55=−255\tan A = -\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}

Step 4: Determine Other Trigonometric Functions

With sin⁡A\sin A, cos⁡A\cos A, and tan⁡A\tan A known, we can find the other trigonometric functions using their reciprocal relationships:

  • Cosecant (csc⁡A\csc A): The reciprocal of sin⁡A\sin A

csc⁡A=1sin⁡A=123=32\csc A = \frac{1}{\sin A} = \frac{1}{\frac{2}{3}} = \frac{3}{2}

  • Secant (sec⁡A\sec A): The reciprocal of cos⁡A\cos A

sec⁡A=1cos⁡A=1−53=−35=−355\sec A = \frac{1}{\cos A} = \frac{1}{-\frac{\sqrt{5}}{3}} = -\frac{3}{\sqrt{5}} = -\frac{3\sqrt{5}}{5}

  • Cotangent (cot⁡A\cot A): The reciprocal of tan⁡A\tan A

cot⁡A=1tan⁡A=1−255=−525=−52\cot A = \frac{1}{\tan A} = \frac{1}{-\frac{2\sqrt{5}}{5}} = -\frac{5}{2\sqrt{5}} = -\frac{\sqrt{5}}{2}

Step 5: Summarize the Results

We have successfully determined the values of all six trigonometric functions of angle A without using a calculator:

  • sin⁡A=23\sin A = \frac{2}{3}
  • cos⁡A=−53\cos A = -\frac{\sqrt{5}}{3}
  • tan⁡A=−255\tan A = -\frac{2\sqrt{5}}{5}
  • csc⁡A=32\csc A = \frac{3}{2}
  • sec⁡A=−355\sec A = -\frac{3\sqrt{5}}{5}
  • cot⁡A=−52\cot A = -\frac{\sqrt{5}}{2}

This comprehensive solution showcases the power of understanding and applying trigonometric identities and relationships. By carefully using the given information and the Pythagorean identity, we were able to find the values of all trigonometric functions without relying on a calculator. This method highlights the importance of a strong foundation in trigonometric principles.

Mastering trigonometric identities and values is essential for success in mathematics and related fields. By understanding the relationships between trigonometric functions and applying fundamental identities, we can solve a wide range of problems efficiently and accurately. This guide has provided a detailed exploration of key concepts and techniques, equipping you with the tools to excel in trigonometry. The ability to manipulate and apply trigonometric identities is a valuable skill that enhances problem-solving capabilities in various mathematical contexts. Continuous practice and application of these concepts will solidify your understanding and mastery of trigonometry.