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Chapter 7: Problem 58

Verify that each trigonometric equation is an identity. $$\frac{\sin ^{4} \alpha-\cos ^{4} \alpha}{\sin ^{2} \alpha-\cos ^{2} \alpha}=1$$

Short Answer

Expert verified
\frac{\sin^4 \alpha - \cos^4 \alpha}{\sin^2 \alpha - \cos^2 \alpha} = 1.

Step by step solution

01

- Simplify the numerator

Identify the numerator \( \sin^4 \alpha - \cos^4 \alpha \). Notice it is a difference of squares, which can be factored as \( (\sin^2 \alpha)^2 - (\cos^2 \alpha)^2 = (\sin^2 \alpha - \cos^2 \alpha)(\sin^2 \alpha + \cos^2 \alpha) \).
02

- Simplify the denominator

The denominator is already \( \sin^2 \alpha - \cos^2 \alpha \), which matches one of the factors in the numerator.
03

- Cancel common terms

Cancel the common factor \( \sin^2 \alpha - \cos^2 \alpha \) from the numerator and denominator. We are left with \( \sin^2 \alpha + \cos^2 \alpha \).
04

- Apply trigonometric identity

Recall the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \).
05

- Conclude the identity

After simplification, we are left with 1, so the given equation \( \frac{\sin^4 \alpha - \cos^4 \alpha}{\sin^2 \alpha - \cos^2 \alpha} \) simplifies to 1.

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Key Concepts

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

Difference of Squares
The difference of squares is a mathematical expression of the form \( a^2 - b^2 \). This can be factored into \( (a - b)(a + b) \). In our exercise, the numerator \( \sin^4 \alpha - \cos^4 \alpha \) is a difference of squares. By recognizing this, we can rewrite it as \( (\sin^2 \alpha)^2 - (\cos^2 \alpha)^2 \) and then factor it to \( (\sin^2 \alpha - \cos^2 \alpha)(\sin^2 \alpha + \cos^2 \alpha) \). This step is crucial because it simplifies the expression and reveals common factors that can be canceled, making the process of solving trigonometric identities easier.
Pythagorean Identity
The Pythagorean identity is one of the fundamental identities in trigonometry. It states that \( \sin^2 \alpha + \cos^2 \alpha = 1 \). This identity is derived from the Pythagorean theorem applied to a unit circle. In the exercise, once we have factored and simplified the expression, we end up with this identity. Recognizing \( \sin^2 \alpha + \cos^2 \alpha \) allows us to further simplify our expression to 1. Knowing this identity is key to verifying and simplifying many trigonometric expressions and identities.
Simplifying Trigonometric Expressions
Simplifying trigonometric expressions often involves using algebraic techniques and known identities. Here are the steps in our exercise:
1. Identify patterns such as the difference of squares.
2. Factor where possible. For example, \( \sin^4 \alpha - \cos^4 \alpha \) can be factored into \( (\sin^2 \alpha - \cos^2 \alpha)(\sin^2 \alpha + \cos^2 \alpha) \).
3. Simplify the expressions by canceling common factors. In the exercise, \( \sin^2 \alpha - \cos^2 \alpha \) in both the numerator and denominator can be canceled out.
4. Use known identities like the Pythagorean identity to further simplify the expression.
By following these steps, you can break down complex trigonometric problems into simpler parts, making them easier to handle.

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Most popular questions from this chapter

The model $$ 0.342 D \cos \theta+h \cos ^{2} \theta=\frac{16 D^{2}}{V_{0}^{2}} $$ is used to reconstruct accidents in which a vehicle vaults into the air after hitting an obstruction. \(V_{0}\) is velocity in feet per second of the vehicle when it hits the obstruction, \(D\) is distance (in feet) from the obstruction to the landing point, and \(h\) is the difference in height (in feet) between landing point and takeoff point. Angle \(\theta\) is the takeoff angle, the angle between the horizontal and the path of the vehicle. Find \(\theta\) to the nearest degree if \(V_{0}=60, D=80,\) and \(h=2\)

Verify that each equation is an identity. $$\frac{2}{1+\cos x}-\tan ^{2} \frac{x}{2}=1$$

(This discussion applies to functions of both angles and real numbers.) Consider the following. $$\begin{aligned}\cos (&\left.180^{\circ}-\theta\right) \\\&=\cos 180^{\circ} \cos \theta+\sin 180^{\circ} \sin \theta \\\&=(-1) \cos \theta+(0) \sin \theta \\\&=-\cos \theta\end{aligned}$$ \(\cos \left(180^{\circ}-\theta\right)=-\cos \theta\) is an example of a reduction formula, which is an identity that reduces a function of a quadrantal angle plus or minus \(\theta\) to a function of \(\theta\) alone. Another example of a reduction formula is \(\cos \left(270^{\circ}+\theta\right)=\sin \theta\) Here is an interesting method for quickly determining a reduction formula for a trigonometric function \(f\) of the form \(f(Q \pm \theta),\) where \(Q\) is a quadrantal angle. There are two cases to consider, and in each case, think of \(\boldsymbol{\theta}\) as a small positive angle in order to determine the quadrant in which \(Q \pm \theta\) will lie. Case 1 Suppose that \(Q\) is a quadrantal angle whose terminal side lies along the \(x\) -axis. Determine the quadrant in which \(Q \pm \theta\) will lie for a small positive angle \(\theta .\) If the given function \(f\) is positive in that quadrant, use a \(+\) sign on the reduced form. If \(f\) is negative in that quadrant, use a - sign. The reduced form will have that \(\operatorname{sign}, f\) as the function, and \(\theta\) as the argument. For example: CAN'T COPY THE GRAPH Case 2 Suppose that \(Q\) is a quadrantal angle whose terminal side lies along the \(y\) -axis. Determine the quadrant in which \(Q \pm \theta\) will lie for a small positive angle \(\theta .\) If the given function \(f\) is positive in that quadrant, use a \(+\) sign on the reduced form. If \(f\) is negative in that quadrant, use a - sign. The reduced form will have that sign, the cofunction of \(f\) as the function, and \(\theta\) as the argument. For example: CAN'T COPY THE GRAPH Use these ideas to write reduction formulas for each of the following. $$\sin \left(180^{\circ}+\theta\right)$$

Verify that each equation is an identity. $$\frac{\sin (s-t)}{\sin t}+\frac{\cos (s-t)}{\cos t}=\frac{\sin s}{\sin t \cos t}$$

Verify that each equation is an identity. $$\frac{1+\cos 2 x}{\sin 2 x}=\cot x$$
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