Problem 138
Question
$$ \frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=\tan \theta $$
Step-by-Step Solution
Verified Answer
The given equation \(\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=\tan \theta\) is confirmed to be correct through the use of trigonometric identities and mathematical manipulation.
1Step 1: Simplify the double angle terms
First, observe the double angle terms in both the numerator and denominator of the left side of the equation. These can be simplified by using the double angle formula for sine and cosine. The formula states: \(sin 2 \theta = 2 sin\theta cos\theta\) and \(cos 2 \theta = cos^2 \theta - sin^2 \theta\). Substitute these into the equation to get: \(\frac{\sin \theta+2\sin \theta\cos \theta}{1+\cos \theta+cos^2 \theta-sin^2 \theta}\)
2Step 2: Simplify the equation
Next, combine like terms and observe that the numerator becomes a factored expression of \(\sin \theta(1+2\cos \theta)\). In the denominator, the term \(1 - sin^2 \theta\) is a Pythagorean identity equivalent to \(cos^2 \theta\). Therefore, the equation can be written as: \(\frac{\sin \theta(1+2\cos \theta)}{2\cos^2\theta+\cos\theta}\)
3Step 3: Eliminate the common factors
In both the numerator and denominator, there is a common factor of \(\cos \theta\). Divide each term by \(\cos^2 \theta\), and one factor of \(\cos\theta\) will be eliminated: \(\frac{\tan\theta(1+2/\cos \theta)}{2+\frac{\cos\theta}{\cos\theta}}\) This simplifies to: \(\frac{\tan \theta(1+2 \sec \theta)}{2+1} = \tan \theta\) At this point, the left side now equals the right side, therefore the original equation is true.
Key Concepts
Double Angle FormulasPythagorean IdentitySimplification in Trigonometry
Double Angle Formulas
Double angle formulas are incredibly useful tools in trigonometry. They allow us to simplify expressions involving angles that are twice as large as the given angle. For the sine of double angle, the formula is \( \sin 2\theta = 2 \sin \theta \cos \theta \). This tells us that the sine of twice an angle is twice the product of the sine and cosine of that angle.
On the other hand, the cosine double angle formula can be presented in three different ways:
These identities are particularly helpful as they can transform complex trigonometric expressions into simpler, more manageable forms.
On the other hand, the cosine double angle formula can be presented in three different ways:
- \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)
- \( \cos 2\theta = 2\cos^2 \theta - 1 \)
- \( \cos 2\theta = 1 - 2\sin^2 \theta \)
These identities are particularly helpful as they can transform complex trigonometric expressions into simpler, more manageable forms.
Pythagorean Identity
The Pythagorean Identity is one of the fundamental identities in trigonometry. It states: \( \sin^2 \theta + \cos^2 \theta = 1 \).
This identity is derived from the Pythagorean theorem and can be transformed into other useful forms, such as \( \sin^2 \theta = 1 - \cos^2 \theta \) and \( \cos^2 \theta = 1 - \sin^2 \theta \).
In our exercise, we used the conversion \( 1 - \sin^2 \theta = \cos^2 \theta \) to simplify the denominator of the expression. This step is often critical in solving trigonometric equations because it helps in converting sine into cosine or vice versa. Knowing how to manipulate this identity allows for easy simplification and transformation of expressions, which plays a crucial role in trigonometric proofs.
This identity is derived from the Pythagorean theorem and can be transformed into other useful forms, such as \( \sin^2 \theta = 1 - \cos^2 \theta \) and \( \cos^2 \theta = 1 - \sin^2 \theta \).
In our exercise, we used the conversion \( 1 - \sin^2 \theta = \cos^2 \theta \) to simplify the denominator of the expression. This step is often critical in solving trigonometric equations because it helps in converting sine into cosine or vice versa. Knowing how to manipulate this identity allows for easy simplification and transformation of expressions, which plays a crucial role in trigonometric proofs.
Simplification in Trigonometry
Simplification is a key aspect of solving trigonometric equations. It involves reducing complex expressions into their simplest form by using identities and properties of trigonometric functions. In this problem, simplification started by recognizing the common factors and using the double angle formulas.
When simplifying, our goal was to make the left-hand side of the equation look like the right-hand side, which was \( \tan \theta \).
By factoring out common elements and canceling similar terms, the equation was eventually refined. Moreover, expression like \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) became useful.
When simplifying, our goal was to make the left-hand side of the equation look like the right-hand side, which was \( \tan \theta \).
By factoring out common elements and canceling similar terms, the equation was eventually refined. Moreover, expression like \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) became useful.
- First, identifying similar terms in the numerator and denominator made it simple to factor them out.
- Then, simplifying with respect to \( \cos \theta \) allowed for easy cancellation leading to convenient expressions.
- Finally, reaching the expression \( \frac{\tan \theta(1+2 \sec \theta)}{2+1} = \tan \theta \) proved the original equation is true.
Other exercises in this chapter
Problem 136
$$ \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}=2 \tan 2 A $$
View solution Problem 137
$$ \cot \left(A+15^{\circ}\right)-\tan \left(A-15^{\circ}\right)=\frac{4 \cos 2 A}{1+2 \sin 2 A} $$
View solution Problem 139
$$ \sin (2 n+1) A \sin A=\sin ^{2}(n+1) A-\sin ^{2} n A $$
View solution Problem 140
$$ \frac{\sin (A+3 B)+\sin (3 A+B)}{\sin 2 A+\sin 2 B}=2 \cos (A+B) \text { . } $$
View solution