pre-calculus, only if you got the time...

[zabije cie]

Thanks for at least clicking on the link.

now onto the problems
P.S. if you could, show some work so i know how u did the problems, its for a test/Quiz make up and i have no clue to what im doing, tutoring isnt helping me :S

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Test 1

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++++1)
Prove the Identity:

(Tan^2 theta) / (1+ Tan^2 theta) = sin^2 theta

++++2)
Solve for theta on the interval [0,2pi)

(A)
(2sin theta) (Cosine theta) = sin theta

(B)
(Sin^2 theta) - (sin theta) = cosine^2 theta

++++3)
Find the EXACT values for the Following:

(A)
tan15 (degrees)

(B)
Sin105 (degrees)

++++4)
Given:
0<Beta<pi/2<alpha<pi with sin beta=24/25 and sin alpha=4/5,
FIND sin (Alpha-Beta)

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Quiz 1
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++++1)

Prove the following identities:

(Tan^2 theta) - (Sin^2 theta) = (Sin^2 theta) * (tan^2 theta)

++++2)

(Cos theta) -1 / 1- (SEC theta) = (Cos theta) +1 / 1+ (SEC theta)

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Quiz 2

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++++1)

Consider the graph of the function y=sinx - sqrt(3) cosx

(A)Find the Amplitude and Period of the graph.
(B)Rewrite the Equation of this graph as a transformed Sin function.

++++2)

To the nearest degree, find the Acute angle between the lines 2x+y=10 and x+2y=10


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Test 2
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++++1)

If Sinx + Bcosx = asin(x+ 3pi/4), Find the values of a and B.

++++2)

Find the Acute angle between the line 2x+3y=6 and its inverse.

++++3)

Solve for theta on [0,2pi)

(Cos theta) - (Cos 2theta) = 1

++++4)

(A)
If Angle theta was any quadrant 2 angle, what quadrant(s) could angle 2theta lie in?

(B)
Given (sin theta)=8/17, where pi/2<theta<pi, find the following:
-B1) Sin2theta
-B2) Cos2theta
-B3) the quadrant containing 2theta

(C)
Simplify the following:

(2SinxCosx)^2 + (Cos^2x - Sin^2x)^2


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I really appriciate if anyone even attempts this stuff :S i will be going in for more tutoring and hopefully will get this stuff. Thanks again guys and you really do help out alot, lots of props for helping me on my last problems...

Thanks again,
-Zabije Cie

[HallsofIvy]

Tan^2 theta) / (1+ Tan^2 theta) = sin^2 theta

Not always easiest but I recommend you change everything into sine and cosine. Tan^2 theta= sin^2 theta/cos^2 theta. The left side becomes
(sin^2 theta/cos^2 theta)/(1+ sin^2 theta/cos^2 theta). Now simplify the fraction by multiplying both numerator and denominator by cos^2 theta. (And remember that sin^2 thet+ cos^2 thet= 1.)


Solve (2sin theta) (Cosine theta) = sin theta
Obvious point- any theta such that sin theta= 0 satisfies this since it makes both sides equal to 0. If sin theta is NOT 0, divide both sides by sin theta to get cos theta= 0. Can you solve sin theta= 0 and cos theta= 0?

Solve (Sin^2 theta) - (sin theta) = cosine^2 theta
Write cos^2 theta as 1- sin^2 theta. Replace sin theta by "x" and solve the quadratic equation for x. Solve sin theta= x to find theta.

Find exact value of tan15 (degrees)
15 = 30/2 so if you know either the "half angle" or "double angle" formulas for sine and cosine, this is easy.

Find exact value of Sin105 (degrees)
105= 90+ 15 degrees so use what you learned in the previous problem, together with a sum formula.

Given:
0<Beta<pi/2<alpha<pi with sin beta=24/25 and sin alpha=4/5,
FIND sin (Alpha-Beta)
You should be able to use sin^2+ cos^2= 1 to find cos beta and cos alpha and then use the fact that sin(a-b)= sin a cos b- cos a sin b.

Prove the following identities:

(Tan^2 theta) - (Sin^2 theta) = (Sin^2 theta) * (tan^2 theta)

++++2)

(Cos theta) -1 / 1- (SEC theta) = (Cos theta) +1 / 1+ (SEC theta)

Look back to the first question.

[HallsofIvy]

Consider the graph of the function y=sinx - sqrt(3) cosx

(A)Find the Amplitude and Period of the graph.
(B)Rewrite the Equation of this graph as a transformed Sin function.
It's probably easier to do (B) first. Asin(a+ b)= Asin(a)cos(b)+ Acos(a)cos(b). Obviously, you want a= x. Now you need to find A and b so that
Acos(b)= 1 and A sin(b)= -sqrt(3). It's very simple to find A!

To the nearest degree, find the Acute angle between the lines 2x+y=10 and x+2y=10
What are the slopes of the lines? The slope is also the tangent of the angle the line makes with the x-axis. Can you get a formula for the tangent of the angle between the lines from that?

If Sinx + Bcosx = asin(x+ 3pi/4), Find the values of a and B.
Again, a sin(x+ b)= a sin x cos b+ a cos x sin b. You need to have
a cos(3pi/4)= 1 and then B= a sin(3pi/4).

Find the Acute angle between the line 2x+3y=6 and its inverse.
What do you mean by the inverse of a line? Do you mean the line which is the graph inverse function to the function having graph 2x+ 3y= 6? If so, the function is y= 2- (2/3)x. What is its inverse? What are the slopes of those two lines? Now use what you learned in the 2nd problem above.