Review the proof of \( \sin (A-B)=\sin A \cos B-\cos A \sin B \) \( \sin (A-B)= \) Step 1: \( =\cos \left(\left(\frac{\pi}{2}-\mu\right)+B\right) \) Step 2: \( =\cos \left(\frac{\pi}{2}-(A-B)\right) \) How must the proof be rearranged for the steps to logically follow each other? Step 3 should be step 1 ? Step 4 should be step 1. Steps 1 and 2 must be switched. Steps 1 and 3 must be switched. Step 3: \( =\cos \left(\frac{\pi}{2}-A\right) \cos B-\sin \left(\frac{\pi}{2}-A\right) \sin B \) Step A: \( =\sin A \cos B-\cos A \sin B \) Mark the and retun Save and Exit Nexd Submit
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Review the proof of cos(A - B) = cosAcosB + sinAsinB. Which of the following complete step 4 of the proof? Step 1: sqrt((cosA - cosB)^2 + (sinA - sinB)^2) = sqrt((cos(A- B) - 1)^2 + (sin(A - B) - 0)^2) Step 2: (cosA - cosB)^2 + (sinA - sinB)^2 = (cos(A - B) - 1)^2 + (sin(A - B) - 0)^2 Step 3: Step 4: - 2cosAcosB - 2sinAsinB = - 2cos(A - B) + 1 Step 5: -2 (cosAcosB + sinAsinB = cos(A - B) Step 6: cosAcosB + sinAsinB = cos(A - B) Options: 1 and 1, 2 and 1, (cosAcosB)^2(sinAsinB)^2 and (cos^2(A - B))(sin^2(A - B)), (cos^2A + sin^2A)(cos^2B + sin^2B) and (cos^2(A - B))(sin^2(A - B))
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This exercise explains why knowing the sines and cosines of acute angles is sufficient to find the sines and cosines of nonacute angles. Refer to the following figure. According to the unit circle definitions, what are the coordinates (in terms of $\theta$ ) of the point $P ?$ (FIGURE CAN NOT COPY) Use your answer in part (a) and the result in Exercise 33 to show (or explain) why $$\cos \left(\theta+\frac{\pi}{2}\right)=-\sin \theta \quad \text { and } \quad \sin \left(\theta+\frac{\pi}{2}\right)=\cos \theta$$ Remark: These two formulas show that if you know the cosine and the sine for a first-quadrant angle $\theta$ then you know the cosine and the sine for the secondquadrant angle $\theta+\pi / 2 .$ (In fact, these formulas are valid when $\theta$ is in any quadrant. Although the figures we used in Exercises 33 and 34 show a first-quadrant angle, the method of proof used applies equally well to the other quadrants.) (c) As examples for the formulas in part (b), use your calculator to verify each of the following equations. (I) $\cos \left(1+\frac{\pi}{2}\right)=-\sin 1$ (II) $\sin \left(1+\frac{\pi}{2}\right)=\cos 1$ (III) $\cos \left(17^{\circ}+90^{\circ}\right)=-\sin 17^{\circ}$ (IV) $\sin \left(17^{\circ}+90^{\circ}\right)=\cos 17^{\circ}$
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