write the expression as a constant sin^2+tan^2+cos^2

[WolfSpice]

I need some help with a study guide before tommorrows calculus exam. I just don't get any of the problems in trigonometric identities and the like. I haven't gotten a lot of work done cause i'm really not sure where to go with them or what to do. I really need help on these 7 problems.

Here's whats bothering me.
1) Write the expression as a constant, a single trigonometric function, or a power of a trig function.

Sin^2(theta)+Tan^2(theta)+Cos^2(theta)

I have no clue how to go about this, but the correct answer I should get is the Sec^2(theta)

2)identify the equation as an identity or not.

(1+cscx)/secX=cosx+cotX

I used identities to get it down to 1+(1/tanx)=cosX+(1/tanX) which seems to work and it's supposed to be an identity, but that formula doesn't prove it yet, how do I further condense it?

3)find the exact value of the expression.

sin(11pi/12)

it's supposed to end up as (2?(6) +1)/6 but I have not the slightest clue how to get there.

4)use the given info to find the exact value.
cosA=1/3, 0<A<(pi/2); SinB=-1/2, (3pi/2)<B<2pi find sin(A-B)

5)find the exact value given A=-4/5, with A in Quadrant 4, tanB=7/24 with B in quadrant 3, and CosC=-5/13 with C in quadrant 2.

find Sin2A.

It comes out to be -24/25, but how do I go about getting that?

6)find the exact value using the half number identity.

cos(-pi/8)

it's supposed to come out as 1/2(?(2+?(2))


7)find the exact value of Y

Y=arcsin(-1/2)
using both the unit circle and the calculator I keep getting -30 degrees (or 11pi/6), but it comes out to be pi/6, do I just go across the x axis because the answer was negative or what?


I know that I didn't get alot of work done on these, but I really don't know where to go with them, please help me on any you can.

 

[galactus]

You have no clue about the first one?. You should know the most popular of the identities, \(\displaystyle sin^{2}x+cos^{2}x=1\).

This makes it \(\displaystyle 1+tan^{2}x=sec^{2}x\)

That's it.

For #3, you could rewrite it as \(\displaystyle sin(\frac{2\pi}{3}+\frac{\pi}{4})\)

which has two you can find on the unit circle. Use the addition formula for sine.

\(\displaystyle sin(u+v)=sin(u)cos(v)+cos(u)sin(v)\)

by letting \(\displaystyle u=\frac{2\pi}{3}, \;\ v=\frac{\pi}{4}\)

#7. What is \(\displaystyle 2\pi-\frac{11\pi}{6}\)?. That is the same thing as \(\displaystyle \frac{-\pi}{6}\) except it falls in the 4th quadrant instead of the 1st.

Note the -1/2 instead of 1/2.

Remember that \(\displaystyle y=rsin{\theta}\)

If y=-1 and r=2, then \(\displaystyle sin^{-1}(\frac{-1}{2})={\theta}\)

Go down the y axis to -1 and the radius is 2 units. In other words, go down -1 and across the positive x axis sqrt(3). Because \(\displaystyle \sqrt{(\sqrt{3})^{2}+(-1)^{2}}=2\). See now?.

 

[BigGlenntheHeavy]

If you are having an exam in Calculus tomorrow, these trig expressions should be second nature to you.

If they are not, you are in the wrong math class.

 

[WolfSpice]

well truthfully it's a pre-cal 2 exam and I missed alot of that chapter, so I needed some help.

thanks galactus for the help!