Problem 79 Without doing any algebraic mani... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

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Chapter 6: Problem 79

Without doing any algebraic manipulations, explain why $$ \left(2 \cos ^{2} \theta-1\right)^{2}+(2 \cos \theta \sin \theta)^{2}=1 $$ for every angle $\theta$.

Short Answer

Expert verified

The given equation \[\left(2 \cos^2 \theta - 1\right)^2 + (2 \cos \theta \sin \theta)^2 = 1\] is true for every angle 胃 because the expressions inside the squares can be represented as $\cos(2\theta)$ and $\sin(2\theta)$, and by the Pythagorean identity $\sin^2(2\theta) + \cos^2(2\theta) = 1$, these expressions will always satisfy the equation for any angle 胃, without needing any algebraic manipulations.

Step by step solution

Recall that this problem involves trigonometric functions. One of the most basic and useful trigonometric identities is the Pythagorean identity involving sine and cosine functions which states that for any angle 胃: \[\sin^2\theta + \cos^2\theta = 1\] #Step 2: Observe the equation#

We are supposed to prove that: \[\left(2 \cos^2 \theta - 1\right)^2 + (2 \cos \theta \sin \theta)^2 = 1\] We'll attempt to see if we can connect this trigonometric equation to the identities we mentioned in step 1, without actually manipulating the given equation algebraically. #Step 3: Identify expressions as trigonometric functions#

Observe that the expressions within the squared terms in the given equation can be represented as trigonometric functions: The expression $2\cos^2\theta - 1$ is equivalent to the double-angle identity for cosine: \[\cos(2\theta) = 2\cos^2\theta - 1\] Similarly, the expression $2\cos\theta\sin\theta$ is equivalent to the double-angle identity for sine: \[\sin(2\theta) = 2\cos\theta\sin\theta\] #Step 4: Rewrite the equation using the double-angle identities#

Now that we have identified that the expressions within the squared terms are equivalent to trigonometric functions for double angles, rewrite the given equation using these identities: \[\left(\cos(2\theta)\right)^2 + \left(\sin(2\theta)\right)^2 = 1\] #Step 5: Utilize the Pythagorean identity#

As we can see, we have squared terms of sine and cosine functions in the equation for some angle. In this case, the angle is $2\theta$. We can now apply the Pythagorean identity $\sin^2u + \cos^2u = 1$ for some angle $u$. Here, let $u = 2\theta$: \[\sin^2(2\theta) + \cos^2(2\theta) = 1\] #Step 6: Conclusion#

From Step 5, we can conclude that the given equation \[\left(2 \cos^2 \theta - 1\right)^2 + (2 \cos \theta \sin \theta)^2 = 1\] is true for every angle 胃 because the expressions inside the squares of the equation represent trigonometric functions and, by their nature and the Pythagorean identity, will always satisfy the equation for any angle 胃.

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91影视

Without doing any algebraic manipulations, explain why $$ \left(2 \cos ^{2} \theta-1\right)^{2}+(2 \cos \theta \sin \theta)^{2}=1 $$ for every angle \(\theta\).

Short Answer

Step by step solution

Recall that this problem involves trigonometric functions. One of the most basic and useful trigonometric identities is the Pythagorean identity involving sine and cosine functions which states that for any angle 胃: \[\sin^2\theta + \cos^2\theta = 1\] #Step 2: Observe the equation#

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