Problem 4
Question
Match with each of the expressions on the right with the equivalent expression on the left. (a) \(\cos ^{2} x\) (i) \(1-\cos ^{2} x\) (b) \(\sin ^{2} x\) (ii) \(\quad 1-\sin ^{2} x\) (c) \(\tan ^{2} x\) (iii) \(\frac{1}{2}-\frac{1}{2} \cos 2 x\) (d) \(\sin x\) (iv) \(-1+\sec ^{2} x\) (e) \(\cos x\) (v) \(\quad 1-\sin ^{2}(x+\pi / 2)\) (f) \(-\cos x\) (vi) \(\sin (x-\pi / 2)\) (vii) \(\frac{1}{2}+\frac{1}{2} \cos 2 x\) (viii) \(\cos (x-\pi / 2)\)
Step-by-Step Solution
Verified Answer
The matched pairs are (a-ii), (b-i), (c-iv), (d-viii), (e-v), (f-vi).
1Step 1: Analyze Left Side Expressions and their Identities
Apply the relevant identities on the left side: (a) \(\cos ^{2} x\), (b) \(\sin ^{2} x\), (c) \(\tan ^{2} x\), (d) \(\sin x\), (e) \(\cos x\), and (f) \(-\cos x\). Using the identities, we can conclude: (a) \(1-\sin ^{2} x\) (b) \(1-\cos ^{2} x\) (c) \(-1+\sec ^{2} x\) (d) \(\cos (x-\pi / 2)\) (e) \(\sin (x+\pi / 2)\) (f) \(-\sin (x+\pi / 2)\)
2Step 2: Matching with Right Side Expressions
Let's now match our results from Step 1 with the right side expressions: (a) \(\cos ^{2} x\) -> (ii) \(1-\sin ^{2} x\), (b) \(\sin ^{2} x\) -> (i) \(1-\cos ^{2} x\), (c) \(\tan ^{2} x\) -> (iv) \(-1+\sec ^{2} x\), (d) \(\sin x\) -> (viii) \(\cos (x-\pi / 2)\), (e) \(\cos x\) -> (v) \(\sin (x+\pi / 2)\), (f) \(-\cos x\) -> (vi) \(-\sin (x+\pi / 2)\)
Key Concepts
Pythagorean IdentityAngle Sum and Difference IdentitiesTrigonometric TransformationFunctions
Pythagorean Identity
The Pythagorean Identity is a fundamental concept in trigonometry. It relates the squares of the sine and cosine functions to the number one. The classic expression of the identity is: \[ \sin^2 x + \cos^2 x = 1 \] This could be re-arranged based on what's needed:
Using this identity helps simplify problems because it allows us to move between different trigonometric expressions without changing their value.
- \( \cos^2 x = 1 - \sin^2 x \)
- \( \sin^2 x = 1 - \cos^2 x \)
Using this identity helps simplify problems because it allows us to move between different trigonometric expressions without changing their value.
Angle Sum and Difference Identities
These identities are also critical in trigonometry. They allow us to find the sine, cosine, or tangent of an angle that is expressed as a sum or difference of two other angles. For example, the cosine of a difference is expressed as:\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \] Similarly, the sine of a difference is:\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \] These can be used to simplify and understand the given functions in the problem like \(\cos(x-\pi/2)\) and \(\sin(x-\pi/2)\). For instance:
- \(\cos(x-\pi/2) = \sin x\)
- \(\sin(x-\pi/2) = -\cos x\)
Trigonometric Transformation
Trigonometric transformations involve altering expressions to simplify equations or expressions. This could mean changing the position of angles or expressing functions in different forms using fundamental identities. For example, changing \(-\cos x\) to \(-\sin(x+\pi/2)\) uses transformations and the angle sum and difference identities.
Transformations often involve shifting angles by \(\frac{\pi}{2}\), \(\pi\), or other convenient values. This shift can lead to simple transformations such as:
Transformations often involve shifting angles by \(\frac{\pi}{2}\), \(\pi\), or other convenient values. This shift can lead to simple transformations such as:
- \(-\cos x\) can also be expressed as \(-\sin(x+\pi/2)\) through transformation.
- \(\sin(x+\pi/2) = \cos x\)
Functions
Functions in trigonometry often refer to sine, cosine, tangent, and their reciprocal functions. Each has specific properties and transforms over different ranges:
- **Sine Function**: Periodic with a period of \(2\pi\), it starts at zero and oscillates between -1 and 1.
- **Cosine Function**: Shares the same period and range limitations as sine but is shifted by \(\frac{\pi}{2}\).
- **Tangent Function**: It has a period of \(\pi\) and ranges from negative to positive infinity, crossing zero at integer multiples of \(\pi\).
Other exercises in this chapter
Problem 4
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