Evaluation trig functions

[chengeto]

\(\displaystyle cos(ArcSec(-\sqrt2+\frac{\pi}{4})\)


My attempt to solution:

\(\displaystyle cos(ArcSec(-\sqrt2+\frac{3.141}{4})\)


Guys is it correct to replace \(\displaystyle \frac{\pi}{4}\) since it is \(\displaystyle \frac{\pi}{4}\approx 0.785\)

 

[soroban]

Hello, chengeto!

Basic rule: never change radians to decimals.
\(\displaystyle \text{Would you recognize }0.5236 \text{ to be approximately }\tfrac{\pi}{6}\;(30^o)\:?\)

There must be a typo . . . As written, it is not workable expression.

\(\displaystyle \cos\bigg[\text{arcsec}\left(\text{-}\sqrt{2}\right)+\frac{\pi}{4}\bigg]\)

\(\displaystyle \text{Using pricinpal values: }\:\text{arcsec}\left(\text{-}\sqrt{2}\right) \:=\:\frac{3\pi}{4}\)

\(\displaystyle \text{So we have: }\:\cos\left(\frac{3\pi}{4} + \frac{\pi}{4}\right) \:=\:\cos(\pi) \;=\;-1\)

 

[chengeto]

Hello, chengeto!

Basic rule: never change radians to decimals.
\(\displaystyle \text{Would you recognize }0.5236 \text{ to be approximately }\tfrac{\pi}{6}\;(30^o)\:?\)

There must be a typo . . . As written, it is not workable expression.

\(\displaystyle \cos\bigg[\text{arcsec}\left(\text{-}\sqrt{2}\right)+\frac{\pi}{4}\bigg]\)

\(\displaystyle \text{Using pricinpal values: }\:\text{arcsec}\left(\text{-}\sqrt{2}\right) \:=\:\frac{3\pi}{4}\)

\(\displaystyle \text{So we have: }\:\cos\left(\frac{3\pi}{4} + \frac{\pi}{4}\right) \:=\:\cos(\pi) \;=\;-1\)

\(\displaystyle \text{But if l have something like: }sin( Arcsec(-7/3)+\frac{\pi}{4}\)


Would l still be able to use the principle values from the unit circle

 

[Deleted member 4993]

How did you get the \(\displaystyle \frac{3\pi}{4}\) ? Did you have to calculate late or you memorize it ?

Use unit circle.If you forgot - go to:

http://en.wikipedia.org/wiki/Unit_circle

 

[chengeto]

How did you get the \(\displaystyle \frac{3\pi}{4})\) ? Did you have to calculate late or you memorize it ?

Use unit circle.If you forgot - go to:

http://en.wikipedia.org/wiki/Unit_circle

I understood that part my problem is that will the unit circle be useful if l had something like:

\(\displaystyle sin( Arcsec(\frac{-7}{3})+\frac{\pi}{4}\)


When l look at the unit circle l can't see anything related to\(\displaystyle \frac{-7}{3}\)

 

[Deleted member 4993]

I understood that part my problem is that will the unit circle be useful if l had something like:

\(\displaystyle sin( Arcsec(\frac{-7}{3})+\frac{\pi}{4}\)


When l look at the unit circle l can't see anything related to\(\displaystyle \frac{-7}{3}\)

Not everything will be in unit circle.

Anyway, your problem seem to have typo - as posted does not make sense to me. Look at it carefully and fix it.

 

[chengeto]

I understood that part my problem is that will the unit circle be useful if l had something like:

\(\displaystyle sin( Arcsec(\frac{-7}{3})+\frac{\pi}{4}\)


When l look at the unit circle l can't see anything related to\(\displaystyle \frac{-7}{3}\)

Not everything will be in unit circle.

Anyway, your problem seem to have typo - as posted does not make sense to me. Look at it carefully and fix it.

\(\displaystyle \ sin(arcsec(\frac{-7}{3})+\frac{\pi}{4})\)

I have a calculus 2 midterm tomorrow and l would like to know if there is a way to solve this type of question without a calculator, especially if the number is not within the unit circle ?