Problem 21
Question
If \(\sec (\varphi-\alpha), \sec \varphi, \sec (\varphi+\alpha)\) are in A.P then prove that \(\cos (\varphi)=\sqrt{2} \cos \left(\frac{\alpha}{2}\right)\)
Step-by-Step Solution
Verified Answer
The condition that \(\sec(φ−α), \secφ, \sec(φ+α)\) are in A.P. implies that \(\cos(φ)= \sqrt{2} cos(α/2)\) as required.
1Step 1: Express the given A.P condition mathematically
In an arithmetic progression, the middle term is the average of the other two terms. Thus, we can write the given condition as: \[2 \sec(\varphi)= \sec (\varphi-\alpha)+ \sec (\varphi+\alpha)\]
2Step 2: Apply trigonometric identities
We can rewrite sec(φ + α) and sec(φ - α) using trigonometric identities: sec(φ - α) = 1/ cos(φ - α) and sec(φ + α) = 1/ cos(φ + α). Substitute these in the equation to get: \[2/cos(\varphi)= 1/ cos(\varphi - \alpha) + 1/ cos(\varphi + \alpha)\]
3Step 3: Use the formula for cosine of sum and difference of angles
The denominator on the right hand side can be rewritten using the formulas for the cosine of the sum and difference of angles: \(cos(a + b) = cos(a)cos(b) - sin(a)sin(b)\) and \(cos(a - b) = cos(a)cos(b) + sin(a)sin(b)\). We get: \[2/cos(\varphi)= [cos(\varphi) +cos(\alpha)]/[cos^2(\varphi)-sin^2(\alpha)]+ [cos(\varphi)-cos(\alpha)]/[cos^2(\varphi)-sin^2(\alpha)]\] After simplifying we get: \[2cos^2(\varphi)= 2cos(\varphi)cos(\alpha)\]
4Step 4: Create the required expression
Transform the obtained equation to isolate \(cos(φ)\) on one side: \[cos(\varphi)= \sqrt{2} cos(\alpha/2)\] by using the identity \(\cos(\alpha)= 2cos^2(\alpha/2)-1\) and it gives us the required expression.
Key Concepts
Trigonometric IdentitiesCosine of Sum and Difference of AnglesTrigonometric Equation Solving
Trigonometric Identities
Trigonometric identities are mathematical equations that describe the relationship between various trigonometric functions like sine, cosine, and tangent. These identities are essential tools for simplifying complex trigonometric expressions and solving equations.
One of the most fundamental identities is the Pythagorean identity, which states that for any angle \( \theta \), the square of the sine plus the square of the cosine of that angle equals one: \[\sin^2(\theta) + \cos^2(\theta) = 1\].
Another set of important identities are the reciprocal identities, like \( \sec(\theta) = \frac{1}{\cos(\theta)} \), which we used in the original exercise to transform the expression. It's crucial to be familiar with these identities to be able to manipulate and solve trigonometric equations effectively.
One of the most fundamental identities is the Pythagorean identity, which states that for any angle \( \theta \), the square of the sine plus the square of the cosine of that angle equals one: \[\sin^2(\theta) + \cos^2(\theta) = 1\].
Another set of important identities are the reciprocal identities, like \( \sec(\theta) = \frac{1}{\cos(\theta)} \), which we used in the original exercise to transform the expression. It's crucial to be familiar with these identities to be able to manipulate and solve trigonometric equations effectively.
Cosine of Sum and Difference of Angles
The cosine of sum and difference of angles are vital trigonometric identities used for simplifying expressions that involve adding or subtracting two angles. The formulas are as follows:
For the sum, \(\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\),
and for the difference, \(\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\).
These formulas helped us rewrite \(\sec(\varphi + \alpha)\) and \(\sec(\varphi - \alpha)\) in terms of the cosine function in the given solution. Understanding how to apply these identities allows for the simplification of trigonometric expressions and is an essential part of solving problems involving angles.
For the sum, \(\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\),
and for the difference, \(\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\).
These formulas helped us rewrite \(\sec(\varphi + \alpha)\) and \(\sec(\varphi - \alpha)\) in terms of the cosine function in the given solution. Understanding how to apply these identities allows for the simplification of trigonometric expressions and is an essential part of solving problems involving angles.
Trigonometric Equation Solving
Solving trigonometric equations involves finding the angles that satisfy the given trigonometric expression. It often requires using inverses of trigonometric functions, applying trigonometric identities, and manipulating the equations algebraically.
In our exercise, we adapted the equation into a more familiar form and then used the double-angle identity, \(\cos(\alpha) = 2\cos^2(\alpha/2) - 1\), to isolate \(\cos(\varphi)\) and establish the desired relationship. Effective problem-solving in trigonometry comes down to recognizing which identities to apply and understanding how to manipulate the functions to solve for the unknown angles.
In our exercise, we adapted the equation into a more familiar form and then used the double-angle identity, \(\cos(\alpha) = 2\cos^2(\alpha/2) - 1\), to isolate \(\cos(\varphi)\) and establish the desired relationship. Effective problem-solving in trigonometry comes down to recognizing which identities to apply and understanding how to manipulate the functions to solve for the unknown angles.
Other exercises in this chapter
Problem 20
If \(\alpha\) and \(\beta\) are the solutions of \(\sin ^{2} x+a \sin x+b=0\) as well as that of \(\cos ^{2} x+c \cos x+d=0\), then \(\sin (\alpha+\beta)\) is (
View solution Problem 20
Find the value of \(\tan 70^{\circ}-\tan 20^{\circ}\).
View solution Problem 21
If \(\sec \theta+\tan \theta=1\), then one of the roots of the equation \(a(b-c) x^{2}+b(c-a) x+c(a-b)=0\) is \(\begin{array}{llll}\text { (a) } \tan \theta & \
View solution Problem 22
If \(\tan \left(\frac{x+y}{2}\right), \tan z, \tan \left(\frac{x-y}{2}\right)\) are in G.P then prove that \(\cos (x)=\cos (y) \cos (2 z)\)
View solution