Respuesta :
Answer:
[tex]\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}[/tex]
Step-by-step explanation:
I'm going to use the following identity to help with the difference inside the tangent function there:
[tex]\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}[/tex]
Let [tex]a=\sin^{-1}(\frac{2}{3})[/tex].
With some restriction on [tex]a[/tex] this means:
[tex]\sin(a)=\frac{2}{3}[/tex]
We need to find [tex]\tan(a)[/tex].
[tex]\sin^2(a)+\cos^2(a)=1[/tex] is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.
[tex](\frac{2}{3})^2+\cos^2(a)=1[/tex]
[tex]\frac{4}{9}+\cos^2(a)=1[/tex]
Subtract 4/9 on both sides:
[tex]\cos^2(a)=\frac{5}{9}[/tex]
Take the square root of both sides:
[tex]\cos(a)=\pm \sqrt{\frac{5}{9}}[/tex]
[tex]\cos(a)=\pm \frac{\sqrt{5}}{3}[/tex]
The cosine value is positive because [tex]a[/tex] is a number between [tex]-\frac{\pi}{2}[/tex] and [tex]\frac{\pi}{2}[/tex] because that is the restriction on sine inverse.
So we have [tex]\cos(a)=\frac{\sqrt{5}}{3}[/tex].
This means that [tex]\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}[/tex].
Multiplying numerator and denominator by 3 gives us:
[tex]\tan(a)=\frac{2}{\sqrt{5}}[/tex]
Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:
[tex]\tan(a)=\frac{2\sqrt{5}}{5}[/tex]
Let's continue on to letting [tex]b=\cos^{-1}(\frac{1}{7})[/tex].
Let's go ahead and say what the restrictions on [tex]b[/tex] are.
[tex]b[/tex] is a number in between 0 and [tex]\pi[/tex].
So anyways [tex]b=\cos^{-1}(\frac{1}{7})[/tex] implies [tex]\cos(b)=\frac{1}{7}[/tex].
Let's use the Pythagorean Identity again I mentioned from before to find the sine value of [tex]b[/tex].
[tex]\cos^2(b)+\sin^2(b)=1[/tex]
[tex](\frac{1}{7})^2+\sin^2(b)=1[/tex]
[tex]\frac{1}{49}+\sin^2(b)=1[/tex]
Subtract 1/49 on both sides:
[tex]\sin^2(b)=\frac{48}{49}[/tex]
Take the square root of both sides:
[tex]\sin(b)=\pm \sqrt{\frac{48}{49}[/tex]
[tex]\sin(b)=\pm \frac{\sqrt{48}}{7}[/tex]
[tex]\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}[/tex]
[tex]\sin(b)=\pm \frac{4\sqrt{3}}{7}[/tex]
So since [tex]b[/tex] is a number between [tex]0[/tex] and [tex]\pi[/tex], then sine of this value is positive.
This implies:
[tex]\sin(b)=\frac{4\sqrt{3}}{7}[/tex]
So [tex]\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}[/tex].
Multiplying both top and bottom by 7 gives:
[tex]\frac{4\sqrt{3}}{1}= 4\sqrt{3}[/tex].
Let's put everything back into the first mentioned identity.
[tex]\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}[/tex]
[tex]\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}[/tex]
Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:
[tex]\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}[/tex]
[tex]\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}[/tex]
