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Chapter 12: Problem 103

$$ 4 \sin 2 x-\tan ^{2}\left(x-\frac{\pi}{4}\right)=4 $$

Short Answer

Expert verified
The solution to this equation are the two values \( sin(x)=\sqrt{\frac{5}{9}} \) and \( sin(x)=-\sqrt{\frac{5}{9}} \). The specific solutions will depend on the range of x as given in the problem.

Step by step solution

01

Simplify the equation using known identities

Use the double angle formula for sine, \( sin(2x)=2sin(x)cos(x) \). Rewritten, the equation is \( 8sin(x)cos(x)- \tan^{2}(x-\frac{\pi}{4})=4 \).
02

Substitution to simplify

The equation \( \tan^{2}(x-\frac{\pi}{4}) \) can be rewritten as \( \tan^{2}(x)-1 \), and substituted back into our original equation. The equation changes to \( 8sin(x)cos(x)- \tan^{2}(x)+1=4 \) or \( 8sin(x)cos(x)- \tan^{2}(x)=3 \).
03

Use Pythagorean identity

Recall from the Pythagorean identity that \( sin^{2}(x) + cos^{2}(x) = 1 \), hence \( 8sin(x)cos(x) = 8\sqrt{1 - sin^{2}(x)}cos(x) \) and our equation simplifies to \( 8\sqrt{1-sin^{2}(x)}cos(x)- \tan^{2}(x) = 3 \). At this point, aim to make sin(x) subject of the formula.
04

Further simplifications

Rewrite \( \tan^{2}(x) = \frac{sin^{2}(x)}{cos^{2}(x)} \) and the equation becomes \( 8\sqrt{1-sin^{2}(x)}cos(x)- \frac{sin^{2}(x)}{cos^{2}(x)} = 3 \). Substitute \( cos^{2}(x)=1-sin^{2}(x) \) to get \( 8\sqrt{1-sin^{2}(x)}\sqrt{1-sin^{2}(x)}- sin^{2}(x) = 3 \) which gives a quadratic equation \( 8(1-sin^{2}(x))- sin^{2}(x) = 3 \) after further simplification.
05

Solve for sin(x)

Solve this quadratic equation \( 8 - 9sin^{2}(x) = 3 \) for sin(x). Rearranging gives \( 9sin^{2}(x) = 5 \) and solving for sin(x) gives two possible solutions: \( sin(x)=\sqrt{\frac{5}{9}} \) or \( sin(x)=-\sqrt{\frac{5}{9}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Angle Formula
The double angle formulas in trigonometry are essential for simplifying expressions involving trigonometric functions of twice an angle, such as \(2x\) in our case. One such formula is for the sine of a double angle, expressed as \(\sin(2x) = 2\sin(x)\cos(x)\).

This formula allows the transformation of a more complicated trigonometric expression into products of simpler ones, making calculations and algebraic manipulations more straightforward. In the given exercise, applying the double angle formula converted the original term \(4\sin(2x)\) into \(8\sin(x)\cos(x)\), which is easier to work with and is a stepping stone towards finding the solution.
Pythagorean Identity
The Pythagorean identity is another cornerstone of trigonometry that relates the squares of the sine and cosine of an angle to the number one. It is stated as \(\sin^2(x) + \cos^2(x) = 1\).

This identity derives from the Pythagorean theorem concerning right-angled triangles and can be used to simplify equations by expressing one trigonometric function in terms of another. In the exercise, we used the identity to express \(\cos(x)\) as \(\sqrt{1-\sin^2(x)}\), thus transforming the problem into an equation featuring \(\sin(x)\) alone, paving the way towards a solution where \(\sin(x)\) is the variable of interest.
Quadratic Equation
When an algebraic expression is set to zero and can be rearranged in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, this is known as a quadratic equation. It is a fundamental concept in algebra with widespread applications in various fields of mathematics, including calculus and trigonometry.

To solve quadratic equations, we can utilize factoring, completing the square, the quadratic formula, or, as illustrated in the exercise, by transposing and simplifying terms. The exercise eventually presented a quadratic in \(\sin(x)\), and solving this provided the possible values for \(\sin(x)\), leading to the final solution of the trigonometric equation.

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