# Thread: Trig Identities and finding real numbers

1. ## Trig Identities and finding real numbers

I am currently finishing a review of precal for my cal I class. It has been a little while since i have used this kind of math so i am at a stand still. I have five questions.

1. find and equivalent expression involving only sines and cosines, and then simplify it.

sin^2(x) +sin(x)-1+cos^2(x)

2. same as above except:

sec(x) +sin(x)

3.find sec(x) if tan(x)=3/4 and x is in quadrant one.

4. find all real numbers in the interval [0,2pi] that satisfy the equation:

2sin(x)cos(x)+cos(x)=0

5. same as #4 except:

sin(x)=1-2sin^2(x)

I struggled with these kind of problems when i was taking pre-cal. I would greatly appreciate it if someone could help me solve these and teach me a little more about the subject.

2. Originally Posted by davisjl1
I am currently finishing a review of precal for my cal I class. It has been a little while since i have used this kind of math so i am at a stand still. I have five questions.

1. find and equivalent expression involving only sines and cosines, and then simplify it.

sin^2(x) +sin(x)-1+cos^2(x)
Hint: note that $\displaystyle -1 + \cos^2 x = -(1 - \cos^2 x) = - \sin^2 x$

2. same as above except:

sec(x) +sin(x)
Hint: $\displaystyle \sec x = \frac 1{\cos x}$

3.find sec(x) if tan(x)=3/4 and x is in quadrant one.
Hint: since we are in the first quadrant, the secant is positive. now note that $\displaystyle \sec^2 x = 1 + \tan^2 x$

4. find all real numbers in the interval [0,2pi] that satisfy the equation:

2sin(x)cos(x)+cos(x)=0
Hint: factor out the common cosine, you get

$\displaystyle \cos x (2 \sin x + 1) = 0$

$\displaystyle \Rightarrow \cos x = 0$ or $\displaystyle 2 \sin x + 1 = 0 \implies \sin x = - \frac 12$

now find $\displaystyle x$ ....you should know the values by heart that do this

5. same as #4 except:

sin(x)=1-2sin^2(x)
Hint: note that you have

$\displaystyle 2 \sin^2 x + \sin x - 1 = 0$

this is quadratic in $\displaystyle \sin x$ (if you let $\displaystyle y = \sin x$ you would have $\displaystyle 2y^2 + y - 1 = 0$)

thus we can solve this for $\displaystyle \sin x$ like a quadratic equation. then we can find what $\displaystyle x$ is.

3. Thank You Jhevon,

Ok now i got number one.

Number two i found a common denominator and ended up with sin+cos/sin*cos can this be simplified even more?

Number three i am still in the dark on this one.

Number four are the answers 2pi/3 and 4pi/3