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.)
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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!