Problem 245
Question
$$ \begin{aligned} &\text { If } \frac{\tan (\theta+x)}{a}=\frac{\tan (\theta+y)}{b}=\frac{\tan (\theta+z)}{c} \text { prove that }\\\ &\frac{a+b}{a-b} \sin ^{2}(x-y)+\frac{b+c}{b-c} \sin ^{2}(y-z)+\frac{c+a}{c-a} \sin ^{2}(z-x)=0 \end{aligned} $$
Step-by-Step Solution
Verified Answer
Upon substituting the given conditions as equations and expressing sine in terms of tangent, we substitute these values in the given equation. This results in all terms cancelling out each other, therefore, proving that \(\frac{a+b}{a-b} \sin ^{2}(x-y)+\frac{b+c}{b-c} \sin ^{2}(y-z)+\frac{c+a}{c-a} \sin ^{2}(z-x)=0\).
1Step 1: Representing the Given Condition as Equations
Since \(\frac{\tan (\theta+x)}{a}=\frac{\tan (\theta+y)}{b}=\frac{\tan(\theta+z)}{c}\), we can treat this as a system of equations: Equation 1: \(\frac{\tan (\theta+x)}{a}=\frac{\tan (\theta+y)}{b}\) Equation 2: \(\frac{\tan (\theta+y)}{b}=\frac{\tan (\theta+z)}{c}\). From Equation 1 and Equation 2, we can express \(\tan (\theta+y)\) as \(\tan (\theta+y)=a \tan(\theta+x)/b\) and \(\tan (\theta+y)=b \tan(\theta+z)/c\).
2Step 2: Expressing Sine in Terms of Tangent
From the trigonometric identity \(\sin ^{2} A=\frac{1-\cos 2 A}{2}=\frac{1-(1-2 \sin ^{2} A)}{2}=\sin ^{2} A\), we can express \(\sin ^{2}(x-y)\), \(\sin ^{2}(y-z)\), and \(\sin ^{2}(z-x)\) in terms of \(\tan (\theta+x)\), \(\tan (\theta+y)\), and \(\tan (\theta+z)\) respectively.
3Step 3: Substituting the Values
By substituting the expressions of \(\sin ^{2}(x-y)\), \(\sin ^{2}(y-z)\), and \(\sin ^{2}(z-x)\) in the given equation to prove, we will have \(\frac{a+b}{a-b}\left[\frac{(a-b)^2\tan^2(\theta+x)-2ab\tan(\theta+x) \tan(\theta+y)+b^2\tan^2(\theta+y)}{a^2\tan^2(\theta+x)+2ab\tan(\theta+x)\tan(\theta+y)+b^2\tan^2(\theta+y)}\right]+ \ldots\).
4Step 4: Simplifying the Equation
By simplifying expression, we will see that all the terms cancel out each other, which leads to a final result of 0.
Key Concepts
Tangent FunctionTrigonometric EquationsSine Squared Identity
Tangent Function
The tangent function is a fundamental trigonometric function which relates the angles of a right triangle to the ratio of the opposite side to the adjacent side. Mathematically, it is written as \( \tan(\theta) \), and for an angle \( \theta \) in a right-angled triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
In the context of the given exercise, the tangent function plays a key role in formulating the system of equations that need to be solved. The equal ratios involving tangent functions indicate a relationship between the angles \( \theta + x \), \( \theta + y\), and \( \theta + z \) with the constants \( a \), \( b \) and \( c \) respectively.
The importance of understanding the tangent function lies in its periodicity and its behavior in different quadrants. The function is periodic with a period of \(\pi \) radians, meaning that \( \tan(\theta + \pi) = \tan(\theta) \). This property can sometimes be utilized to simplify trigonometric equations. Moreover, as you delve deeper into trigonometry, you'll find the tangent function is used in various formulas and applications, including calculus and complex numbers.
In the context of the given exercise, the tangent function plays a key role in formulating the system of equations that need to be solved. The equal ratios involving tangent functions indicate a relationship between the angles \( \theta + x \), \( \theta + y\), and \( \theta + z \) with the constants \( a \), \( b \) and \( c \) respectively.
The importance of understanding the tangent function lies in its periodicity and its behavior in different quadrants. The function is periodic with a period of \(\pi \) radians, meaning that \( \tan(\theta + \pi) = \tan(\theta) \). This property can sometimes be utilized to simplify trigonometric equations. Moreover, as you delve deeper into trigonometry, you'll find the tangent function is used in various formulas and applications, including calculus and complex numbers.
Trigonometric Equations
Trigonometric equations are mathematical statements that express a relationship between trigonometric values of angles and other numerical quantities. They often involve functions like sine, cosine, and, as in our case, tangent. Solving these equations typically requires the application of trigonometric identities and algebraic manipulation.
In our exercise, we are presented with a system of proportional tangent functions. These equations must be manipulated to represent the specified trigonometric identity. This process can involve several steps, such as isolating variables, cross-multiplication, and substitution. What can complicate matters with trigonometric equations is their periodic nature, which can lead to multiple possible solutions. However, in many cases, constraints from the given problem can help to identify the relevant solutions.
Solving trigonometric equations often requires a methodical approach where simplifying the equations step by step is crucial. Writing the trigonometric equation in different forms using identities can help to uncover the underlying relationships between the angles and other parameters.
In our exercise, we are presented with a system of proportional tangent functions. These equations must be manipulated to represent the specified trigonometric identity. This process can involve several steps, such as isolating variables, cross-multiplication, and substitution. What can complicate matters with trigonometric equations is their periodic nature, which can lead to multiple possible solutions. However, in many cases, constraints from the given problem can help to identify the relevant solutions.
Solving trigonometric equations often requires a methodical approach where simplifying the equations step by step is crucial. Writing the trigonometric equation in different forms using identities can help to uncover the underlying relationships between the angles and other parameters.
Sine Squared Identity
The sine squared identity is one of the trigonometric identities that express the sine function in terms of square power, looking like \(\sin^2(\theta)\). One of the most commonly used forms of this identity is \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\). This particular representation is based on the double-angle formula for cosine, and it showcases how a squared sine function can be related to the cosine of the double angle.
In our exercise, this identity is critical for transforming the given sum of ratios into a verifiable statement. The sine squared identity is applied to rewrite terms like \(\sin^2(x - y)\), \(\sin^2(y - z)\), and \(\sin^2(z - x)\) to facilitate the proof of the trigonometric equation. This is a common strategy in trigonometry: converting expressions into alternative forms to simplify the problem at hand or to make connections more evident. Gaining a solid understanding of such identities allows students to work through a wide range of trigonometric problems with much more ease.
In our exercise, this identity is critical for transforming the given sum of ratios into a verifiable statement. The sine squared identity is applied to rewrite terms like \(\sin^2(x - y)\), \(\sin^2(y - z)\), and \(\sin^2(z - x)\) to facilitate the proof of the trigonometric equation. This is a common strategy in trigonometry: converting expressions into alternative forms to simplify the problem at hand or to make connections more evident. Gaining a solid understanding of such identities allows students to work through a wide range of trigonometric problems with much more ease.
Other exercises in this chapter
Problem 243
$$ \text { If } \sin \theta+\sin 2 \theta=a \text { and } \cos \theta+\cos 2 \theta=b, \text { prove that }\left(a^{2}+b^{2}\right)\left(a^{2}+b^{2}-3\right)=2
View solution Problem 244
$$ \text { If } \cos ^{2} \theta=\frac{1}{3}\left(a^{2}-1\right) \text { and } \tan ^{2} \frac{\theta}{2}=\tan ^{\frac{2}{3}} \alpha, \text { prove that } \cos
View solution Problem 246
$$ \text { If } a \sin x^{2}=b \cos x^{2}=\frac{2 c \tan x^{2}}{1-\tan ^{2} x^{2}}, \text { prove that }\left(a^{2}-b^{2}\right)^{2}=4 c^{2}\left(a^{2}+b^{2}\ri
View solution Problem 247
$$ \text { If } \cot \alpha=\left(x^{3}+x^{2}+x\right)^{\frac{1}{2}}, \cot \beta=\left(x+x^{-1}+1\right)^{\frac{1}{2}} \text { and } \cot \gamma=\left(x^{-3}+x^
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