Trigonometric Identities Problem

[courtbits]

1) If \(\displaystyle \tan(\pi/4)=1\), find \(\displaystyle \cot(\pi-\pi/4)\).

2) If \(\displaystyle \cot(17^{\circ}) = 3.2709\), find \(\displaystyle \tan(73^{\circ})\)

3) If \(\displaystyle \cot(\theta) = \frac{-9}{2}\) with \(\displaystyle \theta\) in Quadrant II, find \(\displaystyle \sin (\theta)\)

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I really have no idea how to solve any of these problems. I have more problems similar to it, but I thought one of each different type of problem would help me possibly solve others.
I may have more questions relating to how you got a term in between each step, also if you could possible link a website that shows step-by-step or even all the identities that relate to the problem I shown above, that would be glorious!
I know it's a lot, but thanks in advance!

 

[Greg]

1. Use the identities \(\displaystyle \sin(\pi-x)=\sin(x)\) and \(\displaystyle \cos(\pi-x)=-\cos(x)\). Do you know \(\displaystyle \sin\left(\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)\)?

2. Use the identities \(\displaystyle \cos(90^\circ-x)=\sin(x)\) and \(\displaystyle \sin(90^\circ-x)=\cos(x)\).

3. \(\displaystyle \cot(\theta)=-\dfrac92\)

\(\displaystyle \dfrac{\cos(\theta)}{\sin(\theta)}=-\dfrac92\)

\(\displaystyle 2\cos(\theta)=-9\sin(\theta)\)

Square both sides:

\(\displaystyle 4\cos^2(\theta)=81\sin^2(\theta)\)

Use the identity \(\displaystyle \sin^2(x)+\cos^2(x)=1\implies1-\sin^2(x)=\cos^2(x)\).

\(\displaystyle 4(1-\sin^2(\theta))=81\sin^2(\theta)\)

\(\displaystyle 4=85\sin^2(\theta)\)

\(\displaystyle \sin^2(\theta)=\dfrac{4}{85}\)

\(\displaystyle \sin(\theta)=\pm\dfrac{2}{\sqrt{85}}\)

As \(\displaystyle \theta\) is in the second quadrant, we choose the positive root:

\(\displaystyle \sin(\theta)=\dfrac{2}{\sqrt{85}}\)

For a list of identities and related information, see here.

 

[Prove It]

1) If \(\displaystyle \tan(\pi/4)=1\), find \(\displaystyle \cot(\pi-\pi/4)\).

2) If \(\displaystyle \cot(17^{\circ}) = 3.2709\), find \(\displaystyle \tan(73^{\circ})\)

3) If \(\displaystyle \cot(\theta) = \frac{-9}{2}\) with \(\displaystyle \theta\) in Quadrant II, find \(\displaystyle \sin (\theta)\)

---------------------------------------------
I really have no idea how to solve any of these problems. I have more problems similar to it, but I thought one of each different type of problem would help me possibly solve others.
I may have more questions relating to how you got a term in between each step, also if you could possible link a website that shows step-by-step or even all the identities that relate to the problem I shown above, that would be glorious!
I know it's a lot, but thanks in advance!

You should know by symmetry that $\displaystyle \begin{align*} \tan{ \left( \pi - \frac{\pi}{4} \right) } = -\tan{ \left( \frac{\pi}{4} \right) } = -1 \end{align*}$, and so what is $\displaystyle \begin{align*} \cot{ \left( \pi - \frac{\pi}{4} \right) } = \frac{1}{\tan{ \left( \pi - \frac{\pi}{4} \right) } } \end{align*}$?

 

[courtbits]

1. Use the identities \(\displaystyle \sin(\pi-x)=\sin(x)\) and \(\displaystyle \cos(\pi-x)=-\cos(x)\). Do you know \(\displaystyle \sin\left(\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)\)?

2. Use the identities \(\displaystyle \cos(90^\circ-x)=\sin(x)\) and \(\displaystyle \sin(90^\circ-x)=\cos(x)\).

3. \(\displaystyle \cot(\theta)=-\dfrac92\)

\(\displaystyle \dfrac{\cos(\theta)}{\sin(\theta)}=-\dfrac92\)

\(\displaystyle 2\cos(\theta)=-9\sin(\theta)\)

Square both sides:

\(\displaystyle 4\cos^2(\theta)=81\sin^2(\theta)\)

Use the identity \(\displaystyle \sin^2(x)+\cos^2(x)=1\implies1-\sin^2(x)=\cos^2(x)\).

\(\displaystyle 4(1-\sin^2(\theta))=81\sin^2(\theta)\)

\(\displaystyle 4=85\sin^2(\theta)\)

\(\displaystyle \sin^2(\theta)=\dfrac{4}{85}\)

\(\displaystyle \sin(\theta)=\pm\dfrac{2}{\sqrt{85}}\)

As \(\displaystyle \theta\) is in the second quadrant, we choose the positive root:

\(\displaystyle \sin(\theta)=\dfrac{2}{\sqrt{85}}\)

For a list of identities and related information, see here.

Problem 3: Why do we have to square both sides?

 

[Greg]

I squared both sides to get an equation in terms of \(\displaystyle \sin^2(\theta)\) which I could then solve for \(\displaystyle \sin(\theta)\).