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Chapter 5: Problem 49

Verify each identity. $$\frac{\cos (x+h)-\cos x}{h}=\cos x \frac{\cos h-1}{h}-\sin x \frac{\sin h}{h}$$

Short Answer

Expert verified
Yes, by applying trigonometric product-to-sum identities, factoring and dividing by 'h', the right hand side of the equation could be perfectly matched with the left hand side, thus verifying the identity.

Step by step solution

01

Expand the RHS

Expand the right hand side (RHS) of the equation using trigonometric product-to-sum identities: \( \cos a \cos b - \sin a \sin b = \cos (a + b) \). So RHS = \( \cos x \cos h - \cos x - \sin x \sin h \) .
02

Factor out \( \cos x \)

Next, factor out \( \cos x \) from the first two terms on the RHS to get: \( \cos x (\cos h - 1) - \sin x \sin h \).
03

Compare the RHS with LHS

Finally, divide both of these parts by 'h', producing \( \cos x \frac{\cos h -1}{h} - \sin x \frac{\sin h}{h} \) , which exactly matches the left hand side (LHS) of the given equation, hence the identity is verified.

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Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning.I've noticed that for sine, cosine, and tangent, the trig function for the sum of two angles is not equal to that trig function of the first angle plus that trig function of the second angle.

Verify each identity. $$\frac{\sin x-\cos x+1}{\sin x+\cos x-1}=\frac{\sin x+1}{\cos x}$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. To prove a trigonometric identity, I select one side of the equation and transform it until it is the other side of the equation, or I manipulate both sides to a common trigonometric expression.

Determine whether each -statement makes sense or does not make sense, and explain your reasoning. There are similarities and differences between solving \(4 x+1=3\) and \(4 \sin \theta+1=3:\) In the first equation, I need to isolate \(x\) to get the solution. In the trigonometric equation, I need to first isolate \(\sin \theta,\) but then \(I\) must continue to solve for \(\theta\)

Exercises \(116-118\) will help you prepare for the material covered in the next section. In each exercise, use exact values of trigonometric functions to show that the statement is true. Notice that each statement expresses the product of sines and/or cosines as a sum or a difference. $$\sin \pi \cos \frac{\pi}{2}=\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{2}\right)+\sin \left(\pi-\frac{\pi}{2}\right)\right]$$
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