IDNLearn.com is your go-to resource for finding answers to any question you have. Discover in-depth answers to your questions from our community of experienced professionals.
Sagot :
To express [tex]\(\sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] only, we will use the following trigonometric identities:
1. [tex]\(\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex].
First, let's rewrite the given expression using this identity:
[tex]\[ \sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right) = \sin \left(\tan ^{-1}\left(\frac{1}{x}\right) - \tan ^{-1}\left(\frac{1}{y}\right)\right) \][/tex]
Next, use the addition formula for sine:
[tex]\[ \sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b) \][/tex]
Set [tex]\(a = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex] and [tex]\(b = \tan^{-1}\left(\frac{1}{y}\right)\)[/tex]. We need to find [tex]\(\sin(a)\)[/tex], [tex]\(\cos(a)\)[/tex], [tex]\(\sin(b)\)[/tex], and [tex]\(\cos(b)\)[/tex].
For [tex]\(a = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex]:
[tex]\[ \tan(a) = \frac{1}{x} \implies a = \tan^{-1}\left(\frac{1}{x}\right) \][/tex]
[tex]\[ \sin(a) = \frac{\frac{1}{x}}{\sqrt{1+\left(\frac{1}{x}\right)^2}} = \frac{1/x}{\sqrt{1+1/x^2}} = \frac{1}{\sqrt{x^2 + 1}} \][/tex]
[tex]\[ \cos(a) = \frac{1}{\sqrt{1+\left(\frac{1}{x}\right)^2}} = \frac{1}{\sqrt{x^2 + 1}} \][/tex]
For [tex]\(b = \tan^{-1}\left(\frac{1}{y}\right)\)[/tex]:
[tex]\[ \tan(b) = \frac{1}{y} \implies b = \tan^{-1}\left(\frac{1}{y}\right) \][/tex]
[tex]\[ \sin(b) = \frac{\frac{1}{y}}{\sqrt{1+\left(\frac{1}{y}\right)^2}} = \frac{1/y}{\sqrt{1+1/y^2}} = \frac{1}{\sqrt{y^2 + 1}} \][/tex]
[tex]\[ \cos(b) = \frac{1}{\sqrt{1+\left(\frac{1}{y}\right)^2}} = \frac{1}{\sqrt{y^2 + 1}} \][/tex]
Substitute these values into the sine difference formula:
[tex]\[ \sin \left(\tan ^{-1}\left(\frac{1}{x}\right) - \tan ^{-1}\left(\frac{1}{y}\right)\right) = \sin(a) \cos(b) - \cos(a) \sin(b) \][/tex]
[tex]\[ = \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{y^2 + 1}} - \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{y^2 + 1}} \][/tex]
Since both terms are identical, they cancel each other out, simplifying to:
[tex]\[ = 0 \][/tex]
Thus, the expression [tex]\(\sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] simplifies to:
[tex]\[ \boxed{\sin \left(\tan^{-1}\left(\frac{1}{x}\right) - \tan^{-1}\left(\frac{1}{y}\right)\right)} \][/tex]
1. [tex]\(\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex].
First, let's rewrite the given expression using this identity:
[tex]\[ \sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right) = \sin \left(\tan ^{-1}\left(\frac{1}{x}\right) - \tan ^{-1}\left(\frac{1}{y}\right)\right) \][/tex]
Next, use the addition formula for sine:
[tex]\[ \sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b) \][/tex]
Set [tex]\(a = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex] and [tex]\(b = \tan^{-1}\left(\frac{1}{y}\right)\)[/tex]. We need to find [tex]\(\sin(a)\)[/tex], [tex]\(\cos(a)\)[/tex], [tex]\(\sin(b)\)[/tex], and [tex]\(\cos(b)\)[/tex].
For [tex]\(a = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex]:
[tex]\[ \tan(a) = \frac{1}{x} \implies a = \tan^{-1}\left(\frac{1}{x}\right) \][/tex]
[tex]\[ \sin(a) = \frac{\frac{1}{x}}{\sqrt{1+\left(\frac{1}{x}\right)^2}} = \frac{1/x}{\sqrt{1+1/x^2}} = \frac{1}{\sqrt{x^2 + 1}} \][/tex]
[tex]\[ \cos(a) = \frac{1}{\sqrt{1+\left(\frac{1}{x}\right)^2}} = \frac{1}{\sqrt{x^2 + 1}} \][/tex]
For [tex]\(b = \tan^{-1}\left(\frac{1}{y}\right)\)[/tex]:
[tex]\[ \tan(b) = \frac{1}{y} \implies b = \tan^{-1}\left(\frac{1}{y}\right) \][/tex]
[tex]\[ \sin(b) = \frac{\frac{1}{y}}{\sqrt{1+\left(\frac{1}{y}\right)^2}} = \frac{1/y}{\sqrt{1+1/y^2}} = \frac{1}{\sqrt{y^2 + 1}} \][/tex]
[tex]\[ \cos(b) = \frac{1}{\sqrt{1+\left(\frac{1}{y}\right)^2}} = \frac{1}{\sqrt{y^2 + 1}} \][/tex]
Substitute these values into the sine difference formula:
[tex]\[ \sin \left(\tan ^{-1}\left(\frac{1}{x}\right) - \tan ^{-1}\left(\frac{1}{y}\right)\right) = \sin(a) \cos(b) - \cos(a) \sin(b) \][/tex]
[tex]\[ = \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{y^2 + 1}} - \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{y^2 + 1}} \][/tex]
Since both terms are identical, they cancel each other out, simplifying to:
[tex]\[ = 0 \][/tex]
Thus, the expression [tex]\(\sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] simplifies to:
[tex]\[ \boxed{\sin \left(\tan^{-1}\left(\frac{1}{x}\right) - \tan^{-1}\left(\frac{1}{y}\right)\right)} \][/tex]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.
