Chapter 19: Problem 1827
If \(\cos \theta+\sec \theta=2\) then \(\cos ^{2012} \theta+\sec ^{2012} \theta=\) (a) \(2^{2012}\) (b) \(2^{2013}\) (c) 2 (d) 0
Short Answer
Expert verified
The value of \(\cos ^{2012} \theta + \sec ^{2012} \theta\) is (c) 2.
Step by step solution
01
Rewrite the given equation in terms of cosine
:
The given equation is \(\cos \theta + \sec \theta = 2\). We know that \(\sec \theta = \frac{1}{\cos \theta}\). So, we can rewrite the given equation as:
\(\cos \theta + \frac{1}{\cos \theta} = 2\)
02
Square both sides
:
Now, we will square both sides of the equation to get rid of the fraction. Keep in mind that squaring both sides of an equation may introduce extraneous solutions, so we will verify our answer at the end.
\((\cos \theta + \frac{1}{\cos \theta})^2 = 2^2\)
Expanding the left side, we get:
\((\cos ^2 \theta + 2 + \frac{1}{\cos ^2 \theta}) = 4\)
03
Solve for \(\cos ^2 \theta\)
:
Now we will solve for \(\cos ^2 \theta\) :
\(\cos ^2 \theta + 2 + \frac{1}{\cos ^2 \theta} = 4\)
Rearranging the equation, we get:
\(\cos ^2 \theta - 2 + \frac{1}{\cos ^2 \theta} = 0\)
Now let \(x = \cos ^2 \theta\) , the above equation can be written as:
\(x - 2 + \frac{1}{x} = 0\)
Multiplying both sides by \(x\), we get:
\(x^2 - 2x + 1 = 0\)
The above equation is a quadratic equation and its solution is:
\(x =1\)
So, \(\cos ^2 \theta = 1\)
04
Find \(\cos ^{2012} \theta + \sec ^{2012} \theta\)
:
Now since we have \(\cos ^2 \theta = 1\), raising both sides to the power 1006 (as \(2012 = 2 \times 1006\)) we get:
\((\cos ^2 \theta)^{1006} = 1^{1006}\)
\(\cos ^{2012} \theta = 1\)
Also, we know that \(\sec \theta = \frac{1}{\cos \theta}\), so \(\sec ^2 \theta = \frac{1}{\cos ^2 \theta}\) and then,
\(\sec ^{2012} \theta= \frac{1}{\cos ^{2012} \theta}= \frac{1}{1}= 1\)
Now, we can substitute these into the expression we want to find:
\(\cos ^{2012} \theta + \sec ^{2012} \theta = 1 + 1 = 2\)
Therefore, the answer is (c) 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cosine and Secant Relationship
Understanding the relationship between cosine (\text{cos}) and secant (\text{sec}) functions is critical when dealing with trigonometric equations. Remember that secant is the reciprocal of cosine, which is mathematically represented as \text{sec} \theta = \frac{1}{\cos \theta}.
This reciprocal relationship makes it possible to convert between the two functions. If we have an equation like \(\text{cos} \theta + \text{sec} \theta = 2\), we can easily substitute \text{sec} \theta with \frac{1}{\text{cos} \theta} to simplify and solve for \text{cos} \theta. Using this substitution technique allows us to avoid working directly with secant, which can sometimes complicate the problem-solving process.
This reciprocal relationship makes it possible to convert between the two functions. If we have an equation like \(\text{cos} \theta + \text{sec} \theta = 2\), we can easily substitute \text{sec} \theta with \frac{1}{\text{cos} \theta} to simplify and solve for \text{cos} \theta. Using this substitution technique allows us to avoid working directly with secant, which can sometimes complicate the problem-solving process.
Squaring Trigonometric Identities
When dealing with trigonometric identities, we often need to transform the original equation to make it more solvable. Squaring both sides is a common method used to eliminate radicals or fractions. But, be cautious—this operation can introduce extraneous solutions.
For instance, if we square \(\text{cos} \theta + \frac{1}{\text{cos} \theta} = 2\), we must expand and simplify carefully, using the identity \(\text{cos}^2 \theta + 2 + \frac{1}{\text{cos}^2 \theta} = 4\).
For instance, if we square \(\text{cos} \theta + \frac{1}{\text{cos} \theta} = 2\), we must expand and simplify carefully, using the identity \(\text{cos}^2 \theta + 2 + \frac{1}{\text{cos}^2 \theta} = 4\).
- Remember that when squaring the sum, you should apply the formula \((a+b)^2 = a^2 + 2ab + b^2\) to get the correct expansion.
- Ensure to verify any potential solutions since not all solutions of the squared equation will satisfy the initial equation.
Solving Quadratic Equations in Trigonometry
When you're solving quadratic equations in trigonometry, you're often dealing with equations where the unknown variable is a trigonometric function, such as \(\cos^2 \theta\) or \(\sin^2 \theta\).
After squaring an equation and rearranging it into the familiar quadratic form \(ax^2 + bx + c = 0\), you can proceed as you usually would in algebra. The quadratic formula or factoring are common methods used to solve for \(x\), which in trigonometric problems is usually a squared trig function.
After squaring an equation and rearranging it into the familiar quadratic form \(ax^2 + bx + c = 0\), you can proceed as you usually would in algebra. The quadratic formula or factoring are common methods used to solve for \(x\), which in trigonometric problems is usually a squared trig function.
Remembering Simple Solutions
- Some quadratic trigonometric equations have simple solutions such as \(x = 1\) or \(x = 0\) when they're perfect squares like \(x^2 - 2x + 1 = 0\).
- After solving the quadratic equation, you must interpret the solution in the context of the trigonometric identity involved, often reverting back to the original trigonometric function.