# Math Help - Multiple choice trigonometry help pls?

1. ## Multiple choice trigonometry help pls?

I'm terrible at trig, mind if I get some explanation on the following questions?

1) What is the exact value of sin2(pi/6) - 2sin(pi/6)cos(pi/6) + cos2(-pi/6)

a)2 - SqRt(3)
b)(2- SqRt(3))/2
c)(2+SqRt(3))/2
d)2+SqRt(3)

2) If cos(x)= -sin(x), then sin2(x) is

a)1/4
b)1/2
c)SqRt(2)/2
d)-SqRt(2)/2

3)For the equation 4cos(x) - SqRt(12) = 0, where 0<x<(or equal to)2pi, determine the measure of x.

a)pi/3, 5pi/3
b)2pi/3, 4pi/3
c)5pi/6 , 7pi/6
d)pi/6 , 11pi/6

4)If cos2(x) - 6sin(2x)=0, then one value of x is

a)0 Degrees
b)90 Degrees
c)180 Degrees
d)255 Degrees

5) For the graph of y=2sin(3x), which of the following is not an x-intercept?

a)60 Degrees
b)90 Degrees
c)240 Degrees

6)What is another form of Sin(A+B)sin(A-B)/cos2(A)cos2(B)

a)cos2(A) - cos2(B)
b)Tan2(A) + Tan2(B)
c)Tan2(A) - Tan2(B)
d)sin2(A) - sin2(B)

2. Originally Posted by Blahdkm
I'm terrible at trig, mind if I get some explanation on the following questions?

1) What is the exact value of sin2(pi/6) - 2sin(pi/6)cos(pi/6) + cos2(-pi/6)

a)2 - SqRt(3)
b)(2- SqRt(3))/2
c)(2+SqRt(3))/2
d)2+SqRt(3)

2) If cos(x)= -sin(x), then sin2(x) is

a)1/4
b)1/2
c)SqRt(2)/2
d)-SqRt(2)/2

3)For the equation 4cos(x) - SqRt(12) = 0, where 0<x<(or equal to)2pi, determine the measure of x.

a)pi/3, 5pi/3
b)2pi/3, 4pi/3
c)5pi/6 , 7pi/6
d)pi/6 , 11pi/6

4)If cos2(x) - 6sin(2x)=0, then one value of x is

a)0 Degrees
b)90 Degrees
c)180 Degrees
d)255 Degrees

5) For the graph of y=2sin(3x), which of the following is not an x-intercept?

a)60 Degrees
b)90 Degrees
c)240 Degrees

6)What is another form of Sin(A+B)sin(A-B)/cos2(A)cos2(B)

a)cos2(A) - cos2(B)
b)Tan2(A) + Tan2(B)
c)Tan2(A) - Tan2(B)
d)sin2(A) - sin2(B)
all your questions need to know these

$sin2x=2sin(x)cos(x),cos2(x)=1-2sin^2(x),$ $cos2(x)=2cos^2(x)-1,cos2(x)=cos^2x-sin^2x$

$sin(A+B)=sinAcosB+cosAsinB,sin(A-B)=sinAcosB-cosAsinB$

$cos\left(\frac{\pi}{6}+2n\pi\right)=\frac{\sqrt{3} }{2}$

$sin\frac{\pi}{4}=\frac{1}{\sqrt{2}},cos\frac{\pi}{ 4}=\frac{1}{\sqrt{2}}$

you should know that
lats take

3) $4cosx-\sqrt{12}=0...4cosx=\sqrt{12}....4cosx=\sqrt{4(3)}$

$4cosx= 2\sqrt{(3)}...cosx=\frac{\sqrt{3}}{2}$ what is angle when cosx = squre root of 3 divded by 2 and it is positive

2)when -sinx = cosx , sin and cosx had a defferent signs this happened in the foruth quarter and in the second quarter and they have the same value that mean that the angle is pi/4 right now pi\4 in the second quarter equal ... and pi/4 in the fourth quarter equal afer you find x find sin2(x)

3. hi

1) $\frac{2-\sqrt{3}}{2}$

Fot this one you need to know the 'standard' angles. $sin(\frac{\pi}{6})=\frac{1}{2}$ etc ... These can be found on internet easily, or in any decent calculus book.

2) $\frac{1}{2}$

$cos(x)=-sin(x)$ when they have the same magnitude, which is only for $x = \frac{\pi}{4}$ . So here we get $sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} \, \Rightarrow sin^{2}(\frac{3\pi}{4}) = \frac{1}{2}$

4. You could try to put them in the calculator. If it doesn't work put them in seperatly and check brackets and things.
I don't really know the calculations but you could give it a try.

5. Originally Posted by Blahdkm
5) For the graph of y=2sin(3x), which of the following is not an x-intercept?

a)60 Degrees
b)90 Degrees
c)240 Degrees
The x-intercepts of y = sin x are $x = n\pi$, where n is an integer. The x-intercepts of y = 2 sin x do not change because the only thing that changes in the graph is the amplitude. However, the graph of y = 2 sin 3x is a horizontal shrink by a factor of 1/3, so the x-intercepts are $x = \frac{n\pi}{3}$, where n is an integer. In other words, the x-intercepts are multiples of 60 degrees. The answer is (b) because 90 isn't a multiple of 60.

01

6. Originally Posted by Blahdkm
6)What is another form of Sin(A+B)sin(A-B)/cos2(A)cos2(B)

a)cos2(A) - cos2(B)
b)Tan2(A) + Tan2(B)
c)Tan2(A) - Tan2(B)
d)sin2(A) - sin2(B)
$\frac{\sin (A + B) \sin (A - B)}{\cos^2 A \cos^2 B}$

$= \frac{(\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B)}{\cos^2 A \cos^2 B}$

$= \frac{\sin^2 A \cos^2 B - \cos^2 A \sin^2 B}{\cos^2 A \cos^2 B}$

$= \frac{\sin^2 A \cos^2 B}{\cos^2 A \cos^2 B} - \frac{\cos^2 A \sin^2 B}{\cos^2 A \cos^2 B}$

$= \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B}$

$= \tan^2 A - \tan^2 B$ .