Join the conversation on IDNLearn.com and get the answers you seek from experts. Join our community to receive prompt and reliable responses to your questions from experienced professionals.
Sagot :
To determine which of the given expressions are identities, we'll examine each expression individually to see if it simplifies to a standard trigonometric identity.
### Expression A:
[tex]\[ \tan\left(x - \frac{\pi}{4}\right) = \tan x - 1 \][/tex]
To verify this, we check if both sides simplify to the same expression.
[tex]\[ \tan\left(x - \frac{\pi}{4}\right) = \frac{\tan x - \tan\frac{\pi}{4}}{1 + \tan x \tan\frac{\pi}{4}} = \frac{\tan x - 1}{1 + \tan x \cdot 1} = \frac{\tan x - 1}{1 + \tan x} \][/tex]
Comparing this to [tex]\(\tan x - 1\)[/tex], it's clear that [tex]\(\frac{\tan x - 1}{1 + \tan x} \neq \tan x - 1\)[/tex]. Hence, this is not an identity.
### Expression B:
[tex]\[ \tan x - \tan y = \frac{\sin (x - y)}{\cos x \cos y} \][/tex]
We use the difference of tangents and angle difference formulas:
[tex]\[ \tan x - \tan y = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y} \][/tex]
[tex]\[ \frac{\sin (x - y)}{\cos x \cos y} = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y} \][/tex]
Both sides simplify to the same expression, confirming this is an identity.
### Expression C:
[tex]\[ \cos\left(x + \frac{\pi}{6}\right) = -\sin\left(x - \frac{\pi}{3}\right) \][/tex]
To verify this, let's use the angle addition and subtraction formulas:
[tex]\[ \cos\left(x + \frac{\pi}{6}\right) = \cos x \cos \frac{\pi}{6} - \sin x \sin \frac{\pi}{6} = \cos x \frac{\sqrt{3}}{2} - \sin x \frac{1}{2} \][/tex]
[tex]\[ -\sin\left(x - \frac{\pi}{3}\right) = -(\sin x \cos \frac{\pi}{3} - \cos x \sin \frac{\pi}{3}) = -(\sin x \frac{1}{2} - \cos x \frac{\sqrt{3}}{2}) = \cos x \frac{\sqrt{3}}{2} - \sin x \frac{1}{2} \][/tex]
Both sides simplify to the same expression, confirming this is an identity.
### Expression D:
[tex]\[ 1 - \tan x \tan y = \frac{\sin (x + y)}{\sin x \sin y} \][/tex]
To verify this, let's use the tangent and sine addition formulas:
[tex]\[ \tan x \tan y = \frac{\sin x \sin y}{\cos x \cos y} \][/tex]
[tex]\[ 1 - \tan x \tan y = 1 - \frac{\sin x \sin y}{\cos x \cos y} = \frac{\cos x \cos y - \sin x \sin y}{\cos x \cos y} \][/tex]
[tex]\[ \frac{\sin (x + y)}{\sin x \sin y} = \frac{\sin x \cos y + \cos x \sin y}{\sin x \sin y} \][/tex]
Since these expressions don't simplify to a common form between both sides, this is not an identity.
### Conclusion
From the above checks, the expressions that are identities are:
- Expression B: [tex]\(\tan x - \tan y = \frac{\sin (x - y)}{\cos x \cos y}\)[/tex]
- Expression C: [tex]\(\cos\left(x + \frac{\pi}{6}\right) = -\sin\left(x - \frac{\pi}{3}\right)\)[/tex]
Thus, the identities are [tex]\(\boxed{B \text{ and } C}\)[/tex].
### Expression A:
[tex]\[ \tan\left(x - \frac{\pi}{4}\right) = \tan x - 1 \][/tex]
To verify this, we check if both sides simplify to the same expression.
[tex]\[ \tan\left(x - \frac{\pi}{4}\right) = \frac{\tan x - \tan\frac{\pi}{4}}{1 + \tan x \tan\frac{\pi}{4}} = \frac{\tan x - 1}{1 + \tan x \cdot 1} = \frac{\tan x - 1}{1 + \tan x} \][/tex]
Comparing this to [tex]\(\tan x - 1\)[/tex], it's clear that [tex]\(\frac{\tan x - 1}{1 + \tan x} \neq \tan x - 1\)[/tex]. Hence, this is not an identity.
### Expression B:
[tex]\[ \tan x - \tan y = \frac{\sin (x - y)}{\cos x \cos y} \][/tex]
We use the difference of tangents and angle difference formulas:
[tex]\[ \tan x - \tan y = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y} \][/tex]
[tex]\[ \frac{\sin (x - y)}{\cos x \cos y} = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y} \][/tex]
Both sides simplify to the same expression, confirming this is an identity.
### Expression C:
[tex]\[ \cos\left(x + \frac{\pi}{6}\right) = -\sin\left(x - \frac{\pi}{3}\right) \][/tex]
To verify this, let's use the angle addition and subtraction formulas:
[tex]\[ \cos\left(x + \frac{\pi}{6}\right) = \cos x \cos \frac{\pi}{6} - \sin x \sin \frac{\pi}{6} = \cos x \frac{\sqrt{3}}{2} - \sin x \frac{1}{2} \][/tex]
[tex]\[ -\sin\left(x - \frac{\pi}{3}\right) = -(\sin x \cos \frac{\pi}{3} - \cos x \sin \frac{\pi}{3}) = -(\sin x \frac{1}{2} - \cos x \frac{\sqrt{3}}{2}) = \cos x \frac{\sqrt{3}}{2} - \sin x \frac{1}{2} \][/tex]
Both sides simplify to the same expression, confirming this is an identity.
### Expression D:
[tex]\[ 1 - \tan x \tan y = \frac{\sin (x + y)}{\sin x \sin y} \][/tex]
To verify this, let's use the tangent and sine addition formulas:
[tex]\[ \tan x \tan y = \frac{\sin x \sin y}{\cos x \cos y} \][/tex]
[tex]\[ 1 - \tan x \tan y = 1 - \frac{\sin x \sin y}{\cos x \cos y} = \frac{\cos x \cos y - \sin x \sin y}{\cos x \cos y} \][/tex]
[tex]\[ \frac{\sin (x + y)}{\sin x \sin y} = \frac{\sin x \cos y + \cos x \sin y}{\sin x \sin y} \][/tex]
Since these expressions don't simplify to a common form between both sides, this is not an identity.
### Conclusion
From the above checks, the expressions that are identities are:
- Expression B: [tex]\(\tan x - \tan y = \frac{\sin (x - y)}{\cos x \cos y}\)[/tex]
- Expression C: [tex]\(\cos\left(x + \frac{\pi}{6}\right) = -\sin\left(x - \frac{\pi}{3}\right)\)[/tex]
Thus, the identities are [tex]\(\boxed{B \text{ and } C}\)[/tex].
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.
