Calculate the function value of sin(x) and cos(x).

[Decent]

Calculate the function value of sin(x) and cos(x).

You got: tan(x) = -1/sqrt(5)


Thanks, !

 

[Deleted member 4993]

Calculate the function value of sin(x) and cos(x).

You got: tan(x) = -1/sqrt(5)


Thanks, !

Use

\(\displaystyle \dfrac{1}{cos^2(\theta)} \ = \ sec^2(\theta) \ = \ 1 \ + \ tan^2(\theta)\)

 

[Decent]

Use

\(\displaystyle \dfrac{1}{cos^2(\theta)} \ = \ sec^2(\theta) \ = \ 1 \ + \ tan^2(\theta)\)

I have tried everything, but there is still something wrong. Can you solve it for me here shortly?


Thanks !

 

[Deleted member 4993]

I have tried everything, but there is still something wrong. Can you solve it for me here shortly?


Thanks !

Please share your work/s - so that we can catch your mistake/s and provide relevant correction/s.

 

[pka]

Calculate the function value of sin(x) and cos(x).
You got: tan(x) = -1/sqrt(5)

Let us use \(\displaystyle \theta \) instead of \(\displaystyle x\). Then you have \(\displaystyle \tan(\theta)=\dfrac{-1}{\sqrt5} \).

Because of the negative that means \(\displaystyle -\frac{\pi}{2}<\theta<0\text{ or }\frac{\pi}{2}<\theta<\pi\).
Which ever we use changes the signs on the answers.

In the right triangle the hypotenuses is \(\displaystyle \sqrt6\). WHY?
So \(\displaystyle \sin(\theta)=\mp\dfrac{1}{\sqrt6} \). WHY?

YOU find \(\displaystyle \cos(\theta) \)