[MOVED] Suppose cos(x +y) = 1/4 and cos x cos y = 5/24,

[REGINA]

Suppose cos(x +y) = 1/4 and cos x cos y = 5/24, What is the value of cos(x-y)?

 

[stapel]

Try using the identities included in another recent instance of this exercise.

Thank you.

Eliz.

 

[i'madodo]

Try using the identity, cos(x+y)=cos(x)cos(y)-sin(x)sin(y), substitute it into the first equation

 

[soroban]

Hello, Regina!

I suppose you need a walk-through . . .

Suppose: \(\displaystyle \:[1]\;\cos(x\,+\,y) \:= \:\frac{1}{4}\) and \(\displaystyle \,[2]\;\cos x\cos y \:= \:\frac{5}{24}\)

What is the value of: \(\displaystyle \,\cos(x\,-\,y)\) ?

We have the identities: \(\displaystyle \:\begin{array}{cc}\bf{(a)}\;\cos(x\,+\,y)\:=\:\cos x\cos y \,-\,\sin x\sin y \\ \bf{(b)}\;\cos(x\,-\,y)\:=\:\cos x\cos y\,+\,\sin x\sin y\end{array}\)

From \(\displaystyle \bf{(a)}\): \(\displaystyle \:\cos(x\,+\,y)\:=\:\cos x\cos y\,-\,\sin x\sin y \:=\:\frac{1}{4}\)
. . . . . . . . . . . . . . . . . . . .\(\displaystyle \downarrow\)
Using [2], we have:. . . . . \(\displaystyle \frac{5}{24}\;-\;\sin x\sin y \;=\;\frac{1}{4}\;\;\Rightarrow\;\;\sin x\sin y\,=\,-\frac{1}{24}\;\;[3]\)


From \(\displaystyle \bf{(b)}\), we have:\(\displaystyle \;\;\cos(x\,-\,y)\:=\:\cos x\cos y \,+\,\sin x\sin y\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \downarrow\) . . . . . . \(\displaystyle \downarrow\)
Using [2] and [3]: . \(\displaystyle \cos(x\,-\,y)\;=\;\;\;\,\frac{5}{24}\;\;+\;\;\left(-\frac{1}{24}\right) \;\;=\;\;\frac{4}{24}\)

Therefore: \(\displaystyle \L\:\cos(x\,-\,y)\;=\;\frac{1}{6}\)