# Math Help - Final Review Before Last Test

1. ## Final Review Before Last Test

We are having our last test in Trig on Monday and he gave us a last minute review to do over the weekend. The problem is I missed lots of school (I had mono) and am behind. I got all the notes but it makes no sense to me. Could someone explain to me each of these problems and work them out? The test is going to be way harder than these 8 questions, but we are allowed to use it as an outline. Any help on any of these is greatly appreciated
(I use ^ when raising a power)

1) Solve sin^2x = 3cos^2x [sin(squared)=3cos(squared)x]
Find all answers for one revolution

2) Solve 2sin^2x+3sinx+1 = 0
One Revolution

3) Find the exact value of cos 105 (degrees) using sum and difference formulas (***I will provide if you need these***)

4) Fine the exact value of sin(19π)/12 using sum and difference formulas (that is the best my PC can render pi)

5) Find the exact value of the function tan (u+v) given that sin u = -7/25 and cos v = -4/5 (***Both U and V are in 3rd Quadrant)

6) sin2xsinx = cosx
Find exact solutions for one revolution

7) Find the exact value of sin2u cos2u and tan2u using double angle formulas and given that tan u =3/4 0<u<π/2 (again best pi I can make)

8) Use the sum to product formulas to find the exact value of the expression:
sin (5π)/4 - sin (3π)/4

Thanks again so much and it will greatly help me on Monday's test if I understood these and had correct answers!

Welcome to Math Help Forum!

Since you've posted so many questions here, I'll get you started on each one, and leave you to (hopefully) finish off. If you simply can't see what to do next, come back to us.
We are having our last test in Trig on Monday and he gave us a last minute review to do over the weekend. The problem is I missed lots of school (I had mono) and am behind. I got all the notes but it makes no sense to me. Could someone explain to me each of these problems and work them out? The test is going to be way harder than these 8 questions, but we are allowed to use it as an outline. Any help on any of these is greatly appreciated
(I use ^ when raising a power)

1) Solve sin^2x = 3cos^2x [sin(squared)=3cos(squared)x]
Find all answers for one revolution
Divide both sides by $\cos^2x$, noting that $\tan x = \frac{\sin x}{\cos x}$:
$\tan^2x = 3$
Don't forget the $\pm$ sign when you take the square root.

2) Solve 2sin^2x+3sinx+1 = 0
One Revolution
Factorise:
$(2\sin x +1)(\sin x + 1)=0$
Now solve $\sin x = -\tfrac12$ or $-1$.

3) Find the exact value of cos 105 (degrees) using sum and difference formulas (***I will provide if you need these***)
$\cos 105^o = \cos(60^o+45^o)$
$=\cos 60^o\cos45^o-\sin60^o\sin45^o$
4) Fine the exact value of sin(19π)/12 using sum and difference formulas (that is the best my PC can render pi)
$\frac{19}{12}\pi=\frac{24-5}{12}\pi$
$=\frac{24-(3+2)}{12}\pi$

$=2\pi -\left(\frac14 +\frac16\right)\pi$
Now use
$\sin(2\pi-[A+B]) = -\sin (A+B)$
$=-\sin A\cos B + \cos A \sin B$
5) Find the exact value of the function tan (u+v) given that sin u = -7/25 and cos v = -4/5 (***Both U and V are in 3rd Quadrant)
Use
$\cos^2u=1-\sin^2u$ and $\cos u <0$ in QIII to find $\cos u$

$\sin^2v = 1-\cos^2v$ and $\sin v <0$ in QIII to find $\sin v$
Then find $\tan u$ and $\tan v$ before using the addition formula for $\tan(u+v)$.

6) sin2xsinx = cosx
Find exact solutions for one revolution
$\sin2x\sin x = \cos x$

$\Rightarrow 2\sin x \cos x \sin x = \cos x$

$\Rightarrow \cos x = 0$ or $2\sin^2x = 1$
7) Find the exact value of sin2u cos2u and tan2u using double angle formulas and given that tan u =3/4 0<u<π/2 (again best pi I can make)
If $u$ is in QI, then $\sin u = \tfrac35$ and $\cos u = \tfrac45$ (from a 3-4-5 triangle). Now use the double-angle formulae.

8) Use the sum to product formulas to find the exact value of the expression:
sin (5π)/4 - sin (3π)/4

Thanks again so much and it will greatly help me on Monday's test if I understood these and had correct answers!
$\sin A -\sin B = 2 \cos \tfrac12(A+B)\sin\tfrac12(A-B)$

$\Rightarrow \sin \tfrac{5\pi}{4} -\sin \tfrac{3\pi}{4} = 2 \cos \tfrac12(\tfrac{5\pi}{4}+\tfrac{3\pi}{4})\sin\tfra c12(\tfrac{5\pi}{4}-\tfrac{3\pi}{4})$

$= ...?$