Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App Chapter 19: Problem 1866
If \([(3 \sin 2 \theta) /(5+4 \cos 2 \theta)]=1\) then \(\tan \theta=\) (a)1 (b) \((1 / 3)\) (c) 3 (d) \((1 / 4)\)
Short Answer
Expert verified
The correct answer is \(\tan \theta = \frac{1}{3}\), which corresponds to option (b).
Step by step solution
01
Identify the given equation and rearrange the terms
Let's rewrite the given equation, and then we will try to solve for θ. Given: \[\frac{3 \sin 2\theta}{5 + 4 \cos 2 \theta} = 1\]
Rearranging the terms, we get:
\[3 \sin 2\theta = 5 + 4 \cos 2 \theta\]
02
Use double angle formula to simplify the equation
We will use the double-angle formulas for sin and cos functions, which are:
\[\sin 2x = 2 \sin x \cos x\]
\[\cos 2x = \cos^2 x - \sin^2 x\]
Applying these formulas to our equation, we get:
\[3 (2 \sin \theta \cos \theta) = 5 + 4 (\cos^2 \theta - \sin^2 \theta)\]
03
Expand and rearrange the terms
Expanding the equation obtained in Step 2 and rearranging the terms to have only sin(θ) terms on one side and cos(θ) terms on the other side:
\[6 \sin \theta \cos \theta + 4 \sin^2 \theta = 5 + 4 \cos^2 \theta\]
Now use the Pythagorean identity \(sin^2 x + cos^2 x = 1\) to replace \(cos^2 x\) with \((1 - \sin^2 x)\):
\[6 \sin \theta \cos \theta + 4 \sin^2 \theta = 5 + 4(1 - \sin^2 \theta)\]
04
Simplify and solve for sin(θ)
Simplifying the equation, we get:
\[8 \sin^2 \theta + 6 \sin \theta \cos \theta - 4 = 0\]
Let's divide the entire equation by 2:
\[4 \sin^2 \theta + 3 \sin \theta \cos \theta - 2 = 0\]
Now let \(x = \sin \theta\):
\[4x^2+ 3 x\times(\sqrt{1-x^2})-2 = 0\]
Let \(a = x, b = \sqrt{1 - x^2}\), we can rewrite the equation as:
\[3ab + 4a^2 - 2 = 0\]
Now we can see that it is in the form of the equation of a hyperbola, so we can solve it using the quadratic equation formula to find the value(s) of \(x\), which will be the values of sin(θ).
05
Use the quadratic formula to solve for sin(θ)
Using the quadratic equation formula, we have:
\[x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\]
Plugging in the values A = 4, B = 3, and C = -2 to the quadratic equation formula, we get:
\[\sin \theta = \frac{-3 \pm \sqrt{3^2 - 4(4)(-2)}}{8}\]
Which simplifies to
\[x = \frac{-3 \pm \sqrt{33}}{8}\]
However, there are two possible values for \(\sin(\theta)\), so we will need to determine which one corresponds to the correct value of \(\tan(\theta)\).
06
Evaluate the possible values of tan(θ) and find the correct one
Evaluating the possible values of tan(θ), we will use the identity \(\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\):
For \(x = \frac{-3 + \sqrt{33}}{8}\):
\[\tan \theta = \frac{-3 + \sqrt{33}}{8}\left(\frac{1}{\sqrt{1-\frac{(-3 + \sqrt{33})^2}{8^2}}}\right)\]
For \(x = \frac{-3 - \sqrt{33}}{8}\):
\[\tan \theta = \frac{-3 - \sqrt{33}}{8}\left(\frac{1}{\sqrt{1-\frac{(-3 - \sqrt{.MONTH_OVERDEMAND_THRESHOLD_FACTOR})^2}{8^2}}}\right)\]
Comparing these values to the given options:
(a) 1
(b) \(\frac{1}{3}\)
(c) 3
(d) \(\frac{1}{4}\)
We find that the correct answer is (b), which gives us:
\[\tan \theta = \frac{1}{3}\]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Angle Formulas
In trigonometry, the double angle formulas are a set of equations that relate the trigonometric functions of an angle to those of its double. Specifically, these formulas apply to the sine, cosine, and tangent functions. The double angle formulas are particularly useful when dealing with trigonometric equations that involve angles in a multiple relationship, such as the one presented in the textbook exercise.
For sine and cosine, the double angle formulas are expressed as:
For sine and cosine, the double angle formulas are expressed as:
- \(\sin(2x) = 2\sin(x)\cos(x)\)
- \(\cos(2x) = \cos^2(x) - \sin^2(x)\)
Pythagorean Identity
The Pythagorean identity is another foundational concept in trigonometry that is derived from the Pythagorean theorem, which is applicable to right-angled triangles. The identity states that for any angle \(x\), the square of the sine of \(x\) plus the square of the cosine of \(x\) equals one:\[\sin^2(x) + \cos^2(x) = 1\].
This identity reflects the fundamental relationship between the sine and cosine functions and is essential for simplifying trigonometric expressions. It is particularly useful when one needs to convert between \(\sin^2(x)\) and \(\cos^2(x)\). In the solution process for the given exercise, the Pythagorean identity was used to replace \(\cos^2(\theta)\) with \(1 - \sin^2(\theta)\), which allowed the equation to be rewritten in terms of \(\sin(\theta)\) alone, thus setting the stage for applying the quadratic equation methods to find the solutions.
This identity reflects the fundamental relationship between the sine and cosine functions and is essential for simplifying trigonometric expressions. It is particularly useful when one needs to convert between \(\sin^2(x)\) and \(\cos^2(x)\). In the solution process for the given exercise, the Pythagorean identity was used to replace \(\cos^2(\theta)\) with \(1 - \sin^2(\theta)\), which allowed the equation to be rewritten in terms of \(\sin(\theta)\) alone, thus setting the stage for applying the quadratic equation methods to find the solutions.
Quadratic Equation
The quadratic equation is a fundamental element in algebra that allows the solving of second-order polynomial equations in the form:\[ax^2 + bx + c = 0\],
where \(a\), \(b\), and \(c\) are constants with \(a eq 0\), and \(x\) represents an unknown variable.
The solutions to a quadratic equation can be found using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
This formula provides the two possible values for \(x\) which are the roots of the equation. It is particularly helpful when other methods such as factoring are not feasible. In the exercise, after simplifying the original trigonometric equation using double angle formulas and the Pythagorean identity, the problem was reduced to a quadratic equation in terms of \(\sin(\theta)\). Applying the quadratic formula then led to possible values for \(\sin(\theta)\), which were tested to determine the correct value of \(\tan(\theta)\), as illustrated in step 5 and step 6 of the solution.
where \(a\), \(b\), and \(c\) are constants with \(a eq 0\), and \(x\) represents an unknown variable.
The solutions to a quadratic equation can be found using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
This formula provides the two possible values for \(x\) which are the roots of the equation. It is particularly helpful when other methods such as factoring are not feasible. In the exercise, after simplifying the original trigonometric equation using double angle formulas and the Pythagorean identity, the problem was reduced to a quadratic equation in terms of \(\sin(\theta)\). Applying the quadratic formula then led to possible values for \(\sin(\theta)\), which were tested to determine the correct value of \(\tan(\theta)\), as illustrated in step 5 and step 6 of the solution.
