1. ## Few Trig Questions

I need help with a few questions, if someone could do the problem and show how you did it that would be great.

2sinx -1 = o

sinx=squareroot of 3 - sinx

Tan 105 degrees

Find each value of the trig function
Sin u = 3/5 cos u = 7/25

Thanks.

2. Originally Posted by nyyfn26
I need help with a few questions, if someone could do the problem and show how you did it that would be great.

2sinx -1 = o

sinx=squareroot of 3 - sinx

Tan 105 degrees

Find each value of the trig function
Sin u = 3/5 cos u = 7/25

Thanks.
Hi nyyfn26,

Here's a couple of them.

[1] $\displaystyle 2 \sin x-1=0$

$\displaystyle 2 \sin x = 1$

$\displaystyle \sin x = \frac{1}{2}$

$\displaystyle x=\left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\}$ where $\displaystyle 0 \leq x \leq 2 \pi$

[2] $\displaystyle \sin x = \sqrt{3}-\sin x$

$\displaystyle 2 \sin x=\sqrt{3}$

$\displaystyle \sin x = \frac{\sqrt{3}}{2}$

$\displaystyle x=\left\{\frac{\pi}{3}, \frac{2\pi}{3}\right\}$ where $\displaystyle 0 \le x \le 2 \pi$

3. Originally Posted by masters
Hi nyyfn26,

Here's a couple of them.

[1] $\displaystyle 2 \sin x-1=0$

$\displaystyle 2 \sin x = 1$

$\displaystyle \sin x = \frac{1}{2}$

$\displaystyle x=\left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\}$ where $\displaystyle 0 \leq x \leq 2 \pi$

[2] $\displaystyle \sin x = \sqrt{3}-\sin x$

$\displaystyle 2 \sin x=\sqrt{3}$

$\displaystyle \sin x = \frac{\sqrt{3}}{2}$

$\displaystyle x=\left\{\frac{\pi}{3}, \frac{2\pi}{3}\right\}$ where $\displaystyle 0 \le x \le 2 \pi$
Thank you.

4. And another one,

[3] $\displaystyle \tan 105 = \tan(60+45) = \frac{\tan 60+\tan 45}{1-\tan 60 \tan 45}=\frac{\sqrt{3}+1}{1-\sqrt{3}(1)}$

5. And the last one..... I think you mean to find the other trig ratios given the sine ratio of an angle, and given the cosine of another angle. Is that right?

[4] $\displaystyle \sin u=\frac{3}{5}=\frac{y}{r}$

Use Pythagoras to find x: $\displaystyle 5^2=x^2+3^2$

Then, fill in the remaining ratios (could be in QI or QII).

$\displaystyle \tan u=\frac{y}{x} \: \: \: \cot u = \frac{x}{y}$

$\displaystyle \cos u=\frac{x}{r} \: \: \: \sec u = \frac{x}{r}$

$\displaystyle \sin u=\frac{y}{r} \: \: \: \csc u = \frac{r}{y}$

Do something similar with $\displaystyle \cos u = \frac{7}{25}$. Could be in QI or QIV.

6. Thanks a lot man.

I have one more if you don't mind.

2sin2x-1=0