/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 (a) Beginning with the identity ... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Beginning with the identity \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\) prove that \(\cos 2 \theta=1-2 \sin ^{2} \theta\) (b) Using the result in part (a), prove that \(\sin ^{2} \theta=(1-\cos 2 \theta) / 2\) (c) Derive the formula for \(\sin (s / 2)\) as follows: using the identity in part (b), replace \(\theta\) with \(s / 2,\) and then take square roots.

Short Answer

Expert verified
\(\cos 2\theta = 1 - 2\sin^2 \theta\); \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\); \(\sin(s/2) = \sqrt{\frac{1 - \cos s}{2}}\)."

Step by step solution

01

Expand Identity for Part (a)

Start with the given identity: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\). Recall that \(\cos^2 \theta = 1 - \sin^2 \theta\) and substitute into the equation: \(\cos 2\theta = (1 - \sin^2 \theta) - \sin^2 \theta\).
02

Simplify Result for Part (a)

Combine like terms in the expression: \(\cos 2\theta = 1 - 2\sin^2 \theta\). This proves \(\cos 2\theta = 1 - 2\sin^2 \theta\).
03

Rearrange to Prove Part (b)

Starting from \(\cos 2\theta = 1 - 2\sin^2 \theta\), rearrange to solve for \(\sin^2 \theta\): \(2\sin^2 \theta = 1 - \cos 2\theta\).
04

Final Expression for Part (b)

Divide the equation by 2 to isolate \(\sin^2 \theta\): \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\). This confirms the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\).
05

Apply Part (b) for Part (c)

To find \(\sin(s/2)\), replace \(\theta\) with \(s/2\) in \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\). This gives \(\sin^2(s/2) = \frac{1 - \cos s}{2}\).
06

Solve for \(\sin(s/2)\)

Take the square root of both sides of the equation to solve for \(\sin(s/2)\): \(\sin(s/2) = \sqrt{\frac{1 - \cos s}{2}}\).

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Key Concepts

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

Double Angle Formulas
Double angle formulas are powerful tools in trigonometry, allowing us to express trigonometric functions of double angles in terms of single angles. A commonly used double angle identity is for cosine, given by the formula \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta\). This identity can be further transformed. By substituting \( \cos^2 \theta = 1 - \sin^2 \theta\), we derive \( \cos 2\theta = 1 - 2\sin^2 \theta\). This version is especially useful for problems where simplifying in terms of sine is advantageous.

The double angle identities for sine and tangent are also valuable:
  • \( \sin 2\theta = 2\sin \theta \cos \theta\)
  • \( \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}\)
These identities help in solving equations and proving other trigonometric identities by reducing the complexity of multiple angle expressions.
Trigonometric Substitution
Trigonometric substitution is a clever technique, often used to transform algebraic expressions into trigonometric ones to make them more manageable. This strategy is especially useful in integrals and proofs.

In our original exercise, after rearranging \( \cos 2\theta = 1 - 2\sin^2 \theta\), we substitute to isolate \( \sin^2 \theta\). The equation becomes \( 2\sin^2 \theta = 1 - \cos 2\theta \), which can be divided by 2 to yield \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \).

This rearrangement is crucial as it provides a new perspective on how to represent \( \sin^2 \theta \) using a double angle. Trigonometric substitution forms the backbone of simplifying complex trigonometric expressions and proves indispensible in calculus, especially when dealing with non-trivial integrals or derivatives involving trig functions.
Trigonometric Simplification
Trigonometric simplification involves rewriting expressions in a simpler or more convenient form without changing their values. Using identities and algebraic manipulations, we can simplify trigonometric expressions to make them easier to interpret or solve.

In the given exercise, once the expression for \( \sin^2 \theta \) is formulated as \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \), further simplification occurs when deriving \( \sin(s/2) \). By substituting \( \theta \) with \( s/2 \) in the derived formula, the expression becomes \( \sin^2(s/2) = \frac{1 - \cos s}{2} \).

To find \( \sin(s/2) \) itself, take the square root: \( \sin(s/2) = \sqrt{\frac{1 - \cos s}{2}} \). This trigonometric simplification provides a clearer understanding of the function's relationship to the angle \( s \) and aids in solving more complex trigonometric problems.

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Most popular questions from this chapter

Let \(f(t)=\cos ^{2} t+\cos ^{2}\left(t+\frac{2 \pi}{3}\right)+\cos ^{2}\left(t-\frac{2 \pi}{3}\right)\) (a) Complete the table. (Use a calculator.) $$\begin{array}{lllll} \hline t & 1 & 2 & 3 & 4 \\ \hline f(t) & & & & \\ \hline \end{array}$$ (b) On the basis of your results in part (a), make a conjecture about the function \(f .\) Prove that your conjecture is correct.

Show that arctan \(x+\arctan y=\arctan \frac{x+y}{1-x y}\) when \(x\) and \(y\) are positive and \(x y<1.\)

Find all solutions of the equation $$ \sin 3 x \cos x+\cos 3 x \sin x=\frac{\sqrt{3}}{2} $$ in the interval \(0

(a) The angles of a triangle are \(A=20^{\circ}, B=50^{\circ},\) and \(C=110^{\circ} .\) Use your calculator to compute the sum \(\tan A+\tan B+\tan C\) and then the product \(\tan A \tan B \tan C .\) What do you observe? (b) The angles of a triangle are \(\alpha=\pi / 10, \beta=3 \pi / 10\) and \(\gamma=3 \pi / 5 .\) Use your calculator to compute \(\tan \alpha+\tan \beta+\tan \gamma\) and \(\tan \alpha \tan \beta \tan \gamma\) (c) If triangle \(A B C\) is not a right triangle, prove that \(\tan A+\tan B+\tan C=\tan A \tan B \tan C\)

Solve the equations on the interval \([0,2 \pi]\) as follows. Graph the expression on each side of the equation and then zoom in on the intersection points until you are certain of the first three decimal places in each answer. For instance, for Exercise \(53,\) when you graph the two equations \(y=\cos x\)and \(y=0.623\) on the interval \([0,2 \pi],\) you 'll see that there are two intersection points. The \(x\) -coordinates of these points are roots of the equation \(\cos x=0.623\). $$\sin (\cos x)=\sin x$$

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