Some quick review before the test.

[SimonYeahh]

I asked my Trigonometry teacher for some review before the test and I would like to get some help from you people!

1. Suppose secθ = x and θ is acute
a. Sketch a right triangle in which one side has length x and one angle has measure θ, as above.
b. Write an expression for sinθ in terms of x

2. Find the other trig functions of θ, given sinθ=2/3 and cosθ = √5/3

3. Verify the identity: 1-tan^2t/1+tan^2t = 2cos^2t-1

4. Solve the equation where t is any real number: cos2t = -1/2

Thanks in advance for any help!

edit: I don't need any answers, but any hints or clue in which I should use would be perfect. For example, I'm assuming 3 I'd use the concept of verifying identities?

 

[soroban]

Hello, SimonYeahh!

\(\displaystyle \text{1. Suppose }\,\sec\theta = x\,\text{ and }\,\theta\text{ is acute.}\)
. . \(\displaystyle \text{a. Sketch a right triangle in which one side is }x\text{ and one angle is }\theta.\)
. . \(\displaystyle \text{b. Write an expression for }\sin\theta\text{ in terms of }x.\)

We are given: .\(\displaystyle \sec\theta \:=\:\dfrac{x}{1} \:=\:\dfrac{hyp}{adj}\)

\(\displaystyle \theta\) is in a right triangle with: .\(\displaystyle adj = 1,\:hyp = x.\)

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               1

Pythagorus says: .\(\displaystyle opp = \sqrt{x^2-1}\)

Now you can find \(\displaystyle \sin\theta \,=\,\dfrac{opp}{hyp}\)

\(\displaystyle \text{2. Find the other trig functions of }\theta\text{, given }\sin\theta = \frac{2}{3}\text{ and }\cos\theta = \frac{\sqrt{5}}{3}\)

We are given: .\(\displaystyle \sin\theta = \dfrac{2}{3} = \dfrac{opp}{hyp},\;\;\cos\theta = \dfrac{\sqrt{5}}{3} = \dfrac{adj}{hyp}\)

We have: .\(\displaystyle opp = 2,\;adj = \sqrt{5},\;hyp = 3\)

Now you can find: .\(\displaystyle \tan\theta,\;\cot\theta,\;\sec\theta,\;\csc\theta.\)

\(\displaystyle \text{3. Verify the identity: }\:\dfrac{1-\tan^2t}{1+\tan^2t} \:=\:2\cos^2t-1\)

\(\displaystyle \text{The left side is: }\:\dfrac{1-\tan^2t}{1 + \tan^2t} \;=\;\dfrac{1 - \left(\frac{\sin t}{\cos t}\right)^2}{1 + \left(\frac{\sin t}{\cos t}\right)^2} \;=\; \dfrac{1 - \frac{\sin^2t}{\cos^2t}}{1+\frac{\sin^2t}{\cos^2t}} \)

\(\displaystyle \text{Multiply by }\frac{\cos^2t}{\cos^2t}: \;\;\dfrac{\cos^2t - \sin^2t}{\underbrace{\cos^2t + \sin^2t}_{\text{This is 1}}} \;=\;\cos^2t - \sin^2t\)

Now finish it . . .

\(\displaystyle \text{4. Solve the equation where }t\text{ is any real number: }\:\cos2t \,=\,-\frac{1}{2}\)

\(\displaystyle \cos2t \:=\:-\frac{1}{2} \quad\Rightarrow\quad 2t \:=\:\begin{Bmatrix}\frac{2\pi}{3} + 2\pi n \\ \frac{4\pi}{3} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n\)

\(\displaystyle \text{Therefore: }\:t \:=\:\begin{Bmatrix}\frac{\pi}{3} + \pi n \\ \frac{2\pi}{3} + \pi n\end{Bmatrix}\;\text{ for any integer }n\)