Given the sequence 5,1,3 which term of sequence is -75<br>

Prove that (sec 12A-1)/(sec 6A-1)=tan 12A/tan 3A

adelina 88 [10]

Let $x=3A$ . Recall the following identities,

$\cos^2\theta=\dfrac{1+\cos2\theta}2$

$\sin^2\theta=\dfrac{1-\cos2\theta}2$

$\sin2\theta=2\sin\theta\cos\theta$

Now,

$\dfrac{\sec12A-1}{\sec6A-1}=\dfrac{\sec4x-1}{\sec2x-1}$

$=\dfrac{\cos2x(1-\cos4x)}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x(1-\cos^22x)}=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x\sin^22x}=\dfrac{2\cos2x(1+\cos2x)}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x\sin4x}{\cos4x\sin4x}=\dfrac{4\cos2x\cos^2x\tan4x}{\sin4x}$

$=\dfrac{4\cos2x\cos^2x\tan4x}{2\sin2x\cos2x}=\dfrac{2\cos^2x\tan4x}{\sin2x}$

$=\dfrac{2\cos^2x\tan4x}{2\sin x\cos x}=\dfrac{\cos x\tan4x}{\sin x}$

$=\dfrac{\tan4x}{\frac{\sin x}{\cos x}}=\dfrac{\tan4x}{\tan x}=\dfrac{\tan12A}{\tan3A}$

QED

5 0

2 years ago

A baseball has a 45 cm diameter. what is the volume of the contents of the ball

Dvinal [7]

The volume of a sphere is found from the equation V=4/3πr
r is the radius.
If the radius is 45 cm and pi is about 3.14, you can rewrite and equation as V=4/3(3.14)(45)
Then just solve by multiplying all the terms!
4/3(3.14) = 4.19
4.19(45) = 188.55
So the volume of the ball is 188.55 cm cubed.

3 0

2 years ago

A jet flying at 123 m/s banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass of the jet is 2

aev [14]

Answer:

The magnitude of the necessary lifting force is 7.95N

Step-by-step explanation:

Force = mv^2/r

mass (m) = 2.00105kg, velocity (v) = 123m/s, radius (r) = 3810m

Force = 2.00105×123^2/3810 = 30273.89/3810 = 7.95N

7 0

2 years ago

A principal of $2000 is placed in a savings account at 3% per annum compounded annually. How much is in the account after one ye

natita [175]

Answer:

Step-by-step explanation:

After one year

A=p(1+r/n)^nt

=2000(1+0.03/12)^12*1

=2000(1+0.0025)^12

=2000(1.0025)^12

=2000(1.0304)

=$2060.8

After two-years

A=p(1+r/n)^nt

=2060.8(1+0.03/12)^12*2

=2060.8(1+0.0025)^24

=2060.8(1.0025)^24

=2060.8(1.0618)

=$2188.157

After three years

A=p(1+r/n)^nt

=2188.157(1+0.03/12)^12*3

=2188.157(1+0.0025)^36

=2188.157(1.0025)^36

=2188.157(1.0941)

=$2394.063

8 0

2 years ago

Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular

insens350 [35]

The plane we want to find has general equation

$a(x-2)+b(y-5)+c(z-7)=0$

with $a,b,c$ not equal to 0, and has normal vector

$\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k$

$\vec n$ is perpendicular to both the normal vector of the other plane, which is $4\,\vec\imath+5\,\vec\jmath+6\,\vec k$ , as well as the tangent vector to the line $\vec r(t)$ , which is $\vec\imath+2\,\vec\jmath+9\,\vec k$ .

This means the dot product of $\vec n$ with either vector is 0, giving us

$\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}$

Suppose we fix $c=1$ . Then the system reduces to

$\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}$

and we get

$(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10$

$a+2(-10)=-9\implies a=11$

Then one equation for the plane could be

$\boxed{11(x-2)-10(y-5)+(z-7)=0}$

or in standard form,

$\boxed{11x-10y+z=-21}$

The solution is unique up to non-zero scalar multiplication, which is to say that any equation $(11x-10y+z)k=-21k$ would be a valid answer. For example, suppose we instead let $c=2$ ; then we would have found $a=22$ and $b=-20$ , but clearly dividing both sides of the equation

$22(x-2)-20(y-5)+2(z-7)=0$

by 2 gives the same equation as before.

7 0

2 years ago

Given the sequence 5,1,3 which term of sequence is -75​

Given the sequence 5,1,3 which term of sequence is -75