Two questions about trigonomety

[gusrohar]

Hi!

I am currently studying highschool level math over the summer to prepare for uni. I am stuck on two problems:

1) The angle α, measured in radians, meets the requirement π2≤α≤π. Solve for sinα, if you know that cosα=−2/5.

My immediate instinct is to simply solve for alpha by solving cos^(-1)(-2/5) but this gives me the wrong answer!

2)The triangle △ABC is right angled at C and angle α is at corner A. Solve b=|AC|, if c=|AB|=11, and tanα=3

This one I don't have the answer to in my book but I answered 3.478505426 and I was wondering if that was correct?

 

[HallsofIvy]

Hi!

I am currently studying highschool level math over the summer to prepare for uni. I am stuck on two problems:

1) The angle α, measured in radians, meets the requirement π2≤α≤π.

What? Do you mean \(\displaystyle \pi^2\le \alpha\le \pi\) or \(\displaystyle 2\pi\le \alpha\le \pi\)? Neither of those makes sense because both \(\displaystyle \pi^2\) and \(\displaystyle 2\pi\) are larger than \(\displaystyle \pi\)! I will assume you meant \(\displaystyle \pi\le \alpha\le 2\pi\). If that is not correct, let us know.

Solve for sinα, if you know that cosα=−2/5.

My immediate instinct is to simply solve for alpha by solving cos^(-1)(-2/5) but this gives me the wrong answer!

Yes, because the domain of the principle value of cos(x) is \(\displaystyle \pi/2\le x\le \pi/2\) so just "\(\displaystyle cos^{-1}(-2/5)\) would be between \(\displaystyle -\pi/2\) and \(\displaystyle 0\) not between \(\displaystyle \pi\) and \(\displaystyle 2\pi\). Either by looking at a graph of y= cos(x) or by using the fact that cos(x) is periodic with period \(\displaystyle 2\pi\) (if x is between \(\displaystyle -\pi\) and 0 then \(\displaystyle x+ 2\pi\) is between \(\displaystyle \pi\) and \(\displaystyle 2\pi\) you should see that \(\displaystyle cos(x+ 2\pi)= cos(x)\).

2)The triangle △ABC is right angled at C and angle α is at corner A. Solve b=|AC|, if c=|AB|=11, and tanα=3

This one I don't have the answer to in my book but I answered 3.478505426 and I was wondering if that was correct?

By the Pythagorean theorem, \(\displaystyle |BC|^2= |AB|^2- |AC|^2= 121- b^2\) so that \(\displaystyle tan(\alpha)= \frac{\sqrt{121- b^2}}{b}= 3\) so that you have \(\displaystyle 121- b^2= 9b^2\), \(\displaystyle 10b^2= 121\). \(\displaystyle b= \sqrt{12.1}= 3.478505426\).

Although I would not give 9 decimal places. Since the other numbers are given as integers, b= 3.5 should be good enough.

 

[gusrohar]

What? Do you mean \(\displaystyle \pi^2\le \alpha\le \pi\) or \(\displaystyle 2\pi\le \alpha\le \pi\)? Neither of those makes sense because both \(\displaystyle \pi^2\) and \(\displaystyle 2\pi\) are larger than \(\displaystyle \pi\)! I will assume you meant \(\displaystyle \pi\le \alpha\le 2\pi\). If that is not correct, let us know.

Oops! I meant pi/2≤α≤pi.

So in order to solve this question I need to first solve for alpha = cos^(-1)(-2/5) then add or subtract a period and then take the sine of that number?

 

[Deleted member 4993]

Oops! I meant pi/2≤α≤pi.

So in order to solve this question I need to first solve for alpha = cos^(-1)(-2/5) then add or subtract a period and then take the sine of that number?

I think you are expected to use the condition:

sin²(Θ) = 1- cos²(Θ)

and then look at the constraint to choose the correct "sign".