MHB Trigonometric Identities Problem

[courtbits]

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

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

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

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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 $$\sin(\pi-x)=\sin(x)$$ and $$\cos(\pi-x)=-\cos(x)$$. Do you know $$\sin\left(\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)$$?

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

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

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

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

Square both sides:

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

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

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

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

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

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

As $$\theta$$ is in the second quadrant, we choose the positive root:

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

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

 

[Prove It]

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

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

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

---------------------------------------------
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 $$\sin(\pi-x)=\sin(x)$$ and $$\cos(\pi-x)=-\cos(x)$$. Do you know $$\sin\left(\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)$$?

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

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

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

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

Square both sides:

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

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

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

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

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

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

As $$\theta$$ is in the second quadrant, we choose the positive root:

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

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 $$\sin^2(\theta)$$ which I could then solve for $$\sin(\theta)$$.