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Chapter 4: Problem 42

In Exercises \(37-46,\) use trigonometric identities to transform the left side of the equation into the right side \((0<\theta<\pi / 2) .\) $$(1+\cos \theta)(1-\cos \theta)=\sin ^{2} \theta$$

Short Answer

Expert verified
The given equation is correct. By using the Pythagorean identity correctly, we have successfully transformed the left hand side of the equation into the right hand side which is \(sin^2 \theta\).

Step by step solution

01

Expanding the left hand side

Let's start by expanding our left hand side: \((1+cos \theta )(1-cos \theta)\) which results into \(1 - cos^2 \theta\).
02

Applying the Pythagorean Identity

The core relation for this problem is the Pythagorean Identity: \(sin^2 \theta + cos^2 \theta = 1\). We can solve this identity for \(cos^2 \theta\) to get \(cos^2 \theta = 1 - sin^2 \theta\). Therefore, replacing this in our LHS equation we get \(1 - (1 - sin^2 \theta) = sin^2 \theta\).
03

Simplifying the Equation

With simplifying the equation from step 2, we subtract 1 from both sides which results into \(sin^2 \theta\). Therefore left hand side equals to right hand side. Hence, the equation is verified and the transformation is correct.

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Most popular questions from this chapter

In Exercises 91-96, use a graphing utility to graph the function. \(f(x)\ =\ -3\ +\ arctan(\pi x)\)

AREA In calculus, it is shown that the area of the region bounded by the graphs of \(y=0\), \(y=1/(x^2+1)\), \(x=a\), and \(x=b\) is given by Area = arctan \(b\) - arctan \(a\) (see figure). Find the area for the following values of \(a\) and \(b\). (a) \(a=0\), \(b=1\) (b) \(a=-1\), \(b=1\) (c) \(a=0\), \(b=3\) (d) \(a=-1\), \(b=3\)

TRUE OR FALSE? In Exercises 112-114, determine whether the statement is true or false. Justify your answer. \(\arctan\ x\ =\ \dfrac{\arcsin x}{\arccos x}\)

In Exercises 55-66, find the exact value of the expression. (Hint:Sketch a right triangle.) \(sec(arcsin \frac{4}{5})\)

PROOF Prove each identity. (a) \(\arcsin(-x) =\ -\arcsin\ x\) (b) \(\arctan(-x) =\ -\arctan\ x\) (c) \(\arctan\ x\ +\ arctan \dfrac{1}{x} =\ \dfrac{\pi}{2}\), \(x>0\) (d) \(\arcsin\ x\ +\ arccos\ x\ =\ \dfrac{\pi}{2}\) (e) \(\arcsin\ x\ =\ arctan \dfrac{x}{\sqrt{1-x^2}}\)
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