MATH SOLVE

8 months ago

Q:
# I don't need you to work it out, just theorem(s) i need to reach the answer :)

Accepted Solution

A:

Not sure why such an old question is showing up on my feed...

Anyway, let [tex]x=\tan^{-1}\dfrac43[/tex] and [tex]y=\sin^{-1}\dfrac35[/tex]. Then we want to find the exact value of [tex]\cos(x-y)[/tex].

Use the angle difference identity:

[tex]\cos(x-y)=\cos x\cos y+\sin x\sin y[/tex]

and right away we find [tex]\sin y=\dfrac35[/tex]. By the Pythagorean theorem, we also find [tex]\cos y=\dfrac45[/tex]. (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if [tex]\tan x=\dfrac43[/tex], then (by Pythagorean theorem) [tex]\sec x=\dfrac53[/tex], so [tex]\cos x=\dfrac35[/tex]. And from this, [tex]\sin x=\dfrac45[/tex].

So,

[tex]\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}[/tex]

Anyway, let [tex]x=\tan^{-1}\dfrac43[/tex] and [tex]y=\sin^{-1}\dfrac35[/tex]. Then we want to find the exact value of [tex]\cos(x-y)[/tex].

Use the angle difference identity:

[tex]\cos(x-y)=\cos x\cos y+\sin x\sin y[/tex]

and right away we find [tex]\sin y=\dfrac35[/tex]. By the Pythagorean theorem, we also find [tex]\cos y=\dfrac45[/tex]. (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if [tex]\tan x=\dfrac43[/tex], then (by Pythagorean theorem) [tex]\sec x=\dfrac53[/tex], so [tex]\cos x=\dfrac35[/tex]. And from this, [tex]\sin x=\dfrac45[/tex].

So,

[tex]\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}[/tex]