Trigonometric simplification

[klt643]

Hey- I'm having trouble using the sum and difference identities on this section of my homework. someone help!

the question is:
simplify cos(arcsin 4/5 + pi/4)

pi, of course, being the number/symbol 3.14....

Thanks!
-Katie

 

[stapel]

simplify cos(arcsin 4/5 + pi/4)

Is this "cos(arcsin(4/5) + pi/4)", "cos(arcsin(4/5 + pi/4))", or something else?

Thank you.

Eliz.

 

[klt643]

The second one.

 

[klt643]

Thank you so much! ahh that makes perfect sense now- I just forgot that you could distribute the cosine out. thanks again!

 

[soroban]

Hello, Katie!

You're expected to know the sum formula for cosine:

\(\displaystyle \;\;\;\cos(A\,+\,B)\;=\;\cos(A)\cdot\cos(B)\,-\,\sin(A)\cdot\sin(B)\)

Simplify: \(\displaystyle \,\cos\left[\arcsin\frac{4}{5}\,+\,\frac{\pi}{4}\right]\)

We have: \(\displaystyle \,\cos\left[\arcsin\frac{4}{5}\,+\,\frac{\pi}{4}\right] \;=\;\cos\left(\arcsin\frac{4}{5}\right)\cdot\cos\frac{\pi}{4}\,-\,\sin\left(\arcsin\frac{4}{5}\right)\cdot\sin\frac{\pi}{4}\)

We know all four quantities on the right side:

\(\displaystyle \;\;\cos\left(\arcsin\frac{4}{5}\right) \,=\,\cos\left(\arccos\frac{3}{5}\right)\,=\,\frac{3}{5},\;\;\sin\left(\arcsin\frac{4}{5}\right)\,=\,\frac{4}{5},\;\;\sin\frac{\pi}{4}\,=\,\cos\frac{\pi}{4}\,=\,\frac{\sqrt{2}}{2}\)


Therefore: \(\displaystyle \L\,\cos\left[\arcsin\frac{4}{5}\,+\,\frac{\pi}{4}\right] \;= \;\frac{3}{5}\cdot\frac{\sqrt{2}}{2}\,-\,\frac{4}{5}\cdot\frac{\sqrt{2}}{2}\;=\;-\frac{\sqrt{2}}{10}\)


Edit: Yes, I had misplaced the parentheses; it's corrected now.
\(\displaystyle \;\;\;\)Thanks for the heads-up, skeeter!

 

[skeeter]

well, I have to wave the B.S. flag on this one ...

arcsin(4/5 + pi/4) ??? :shock:

how can that be? ... the domain of the arcsin function is [-1, 1].

the original problem should have been stated as cos[arcsin(4/5) + pi/4], which is what Soroban was "thinking" when he solved it correctly.