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work: HW 1-1.1 and 1.3 Question 24, 1.3.31 Part 3 of 6 n list \( 11 \quad 14 \) on 15 on 16 ion 17 ion 18 tion 19 tion 20 stion 21 stion 22 Use an addition formula to derive the identity. \[ \cos \left(x-\frac{\pi}{2}\right)=\sin x \] Which addition formula should be used? \( \cos (A+B) \) \( \boldsymbol{\operatorname { s i n }}(A+B) \) What is the addition formula for \( \cos (A+B) \) ? A. \( \cos (A+B)=\cos A \cos B-\sin A \sin B \) B. \( \cos (A+B)=\cos A \cos B+\sin A \sin B \) C. \( \cos (A+B)=\cos A \sin B+\boldsymbol{\operatorname { s i n }} A \cos B \) D. \( \cos (A+B)=\sin A \sin B-\cos A \cos B \) E. \( \cos (A+B)=\sin A \cos B-\cos A \sin B \) F. \( \cos (A+B)=\sin A \cos B+\cos A \sin B \) Apply the addition formula for \( \cos (A+B) \) to the left side of the given identity. \[ \cos \left(x-\frac{\pi}{2}\right)=\cos (\square) \cos (\square)-\sin (\square) \sin (\square) \] (Type the terms of your expression in the same order as they appear in the original expression. Type exact answers, using \( \pi \) as needed.)

Ask by Davison Cox. in the United States
Jan 20,2025

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Tutor-Verified Answer

Answer

To derive the identity \(\cos \left(x - \frac{\pi}{2}\right) = \sin x\), use the cosine addition formula: \[ \cos (A + B) = \cos A \cos B - \sin A \sin B \] Apply it to \(\cos \left(x - \frac{\pi}{2}\right)\) by setting \(A = x\) and \(B = -\frac{\pi}{2}\): \[ \cos \left(x - \frac{\pi}{2}\right) = \cos x \cos \left(-\frac{\pi}{2}\right) - \sin x \sin \left(-\frac{\pi}{2}\right) \] Since \(\cos \left(-\frac{\pi}{2}\right) = 0\) and \(\sin \left(-\frac{\pi}{2}\right) = -1\), the expression simplifies to: \[ \cos \left(x - \frac{\pi}{2}\right) = 0 + \sin x = \sin x \] Thus, the identity is proven.

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The Deep Dive

To derive the identity \(\cos \left(x-\frac{\pi}{2}\right)=\sin x\), we should definitely use the addition formula for cosine, which is option A: \(\cos (A+B)=\cos A \cos B-\sin A \sin B\). Here, we can treat \(x\) as \(A\) and \(-\frac{\pi}{2}\) as \(B\). Since we're working with subtraction, we replace \(B\) with \(-\frac{\pi}{2}\), making our equation: \[ \cos \left(x - \frac{\pi}{2}\right) = \cos x \cos \left(-\frac{\pi}{2}\right) - \sin x \sin \left(-\frac{\pi}{2}\right). \] Now we just need to determine \(\cos \left(-\frac{\pi}{2}\right)\) and \(\sin \left(-\frac{\pi}{2}\right)\). We know: \[ \cos \left(-\frac{\pi}{2}\right) = 0, \quad \text{and} \quad \sin \left(-\frac{\pi}{2}\right) = -1. \] Plugging these values back into our equation gives: \[ = \cos x \cdot 0 - \sin x \cdot (-1) = \sin x. \] So the final expression is: \[ \cos \left(x-\frac{\pi}{2}\right)=0 - (-\sin x) = \sin x. \] Thus, confirming our identity!

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