Question

Prove: cos θ - cos θ•sin2 θ = cos3 θ. You must show all work.

Answer 1

\[\cos \theta-\cos \theta \times \sin^2 \theta=\cos^3 \theta\]

Answer 2

@zepdrix

Answer 3

@ganeshie8

Answer 4

Oh it's an exponent 3.
I kept thinking it was a (3theta) lol

Answer 5

That would make life complicated XD

Answer 6

\[\Large\rm \color{orangered}{\cos \theta}-\color{orangered}{\cos \theta} \sin ^2\theta\quad=\quad\cos^3\theta\]Let's factor a cos(theta) out of each term.

Answer 7

Get's us here, yes?
\[\Large\rm \color{orangered}{\cos \theta}\left(1-\sin ^2\theta\right)\quad=\quad\cos^3\theta\]

Answer 8

Ok, I'm with you

Answer 9

Recall your Pythagorean Identity for sine and cosine:\[\Large\rm \sin^2\theta+\cos^2\theta=1\]If we subtract sin^2(theta) from each side, we get,\[\Large\rm \color{royalblue}{\cos^2\theta=1-\sin^2\theta}\]


Hmmmmmm, can we apply that to our problem somehow?\[\Large\rm \cos \theta\left(\color{royalblue}{1-\sin ^2\theta}\right)\quad=\quad\cos^3\theta\]

Answer 10

replace the 1 with sin^2+cos^2, then we can cancel out the sin^2, and when we distribute, we'll have cos^3

Answer 11

Yayy!

Answer 12

I think

Answer 13

\[\Large\rm \cos \theta\left(\color{royalblue}{\cos^2\theta}\right)\quad=\quad\cos^3\theta\]

Answer 14

Sorry I ran off for a sec c:
Yayyy good job!

Answer 15

Haha it's totally fine! Could you maybe help me with one more? I think it's relatively short

Answer 16

Sure.

Answer 17

f(x) = 3x^2 + 12x + 16
g(x) = 2 sin(2x - π) + 4


Using complete sentences, explain how to find the minimum value for each function and determine which function has the smallest minimum y-value.

Answer 18

Thanks!

Answer 19

If it were me, I'd just graph them both using tech like geogebra and look for the minimum. In and out in a minute. But that isn't what they want from me

Answer 20

\[\Large\rm f(x)=3x^2+12x+16 \]The leading term (largest term) is a square.
Second power tells us that this will be a parabola.
I guess we would want to put this into vertex form in order to see the vertex (lowest point) easier.

Answer 21

vertex form issssssss?

Answer 22

\[\Large\rm f(x)=a(x-h)^2+k\]See how the x is written in terms of a perfect square?
The vertex (lowest point) will be located at (h,k).
So the minimum value will be k.

Answer 23

\[\Large\rm f(x)=3x^2+12x+16\]Factor a 3 out of the first two terms,\[\Large\rm f(x)=3(\color{green}{x^2+4x})+16\]Now our goal is to try and complete the square on these green terms.

When you have,
\(\Large\rm x^2+bx\)
To complete the square,
You take half of your b coefficient, and square it. \(\Large\rm \left(\frac{b}{2}\right)^2\)
That is the value that will complete the square.

Answer 24

That's still 4

Answer 25

So in our case, the b coefficient is 4.
Half of 4 is 2, 2 squared is 4.\[\Large\rm f(x)=3(\color{green}{x^2+4x+4})+16\]So 4 is the value that will complete the square for us.
But we can't just add 4 willy nilly like this.
We have to also subtract 4 to keep the equation balanced.\[\Large\rm f(x)=3(\color{green}{x^2+4x+4}-4)+16\]We don't want this black 4 inside the parenthesis though.
It isn't part of our perfect square.
So we'll factor it out, multiplying by the 3.

Answer 26

So it comes out like this, 3 times -4
\[\Large\rm f(x)=3(\color{green}{x^2+4x+4})-12+16\]

Answer 27

Does that mean the minimum is 4?

Answer 28

Yes very good!
We can write our x stuff as a perfect square now, but that doesn't really matter.\[\Large\rm f(x)=3(\color{green}{x+2})^2-12+16\]\[\Large\rm f(x)=3(\color{green}{x+2})^2+4\]

Answer 29

Thank you!!

Answer 30

Remember that \(\Large\rm y=x^2\) is a parabola opening upwards.
So we have the same thing going on with our f(x) except the vertex point has been shifted 2 to the right, and 4 up.

Answer 31

ok :)

Answer 32

\[\Large\rm g(x)=2\sin(2x-π)+4\]So this one... hmm let's see.
So our general form:
\[\Large\rm g(x)=A \sin\left[B(x-C)\right]+D\]

Where the function will have an amplitude of \(\Large\rm A\),
A period of \(\Large\rm \frac{2\pi}{B}\),
is shifted horizontally by \(\Large\rm C\),
and is shifted vertically by \(\Large\rm D\).

Answer 33

Let's ignore the period and horizontal shift for now.
Those won't affect the height of the function at all.

Answer 34

So the minimum is 4?

Answer 35

Or is that the Max?

Answer 36

Here is our normal sine curve.
It goes up to 1, down to negative 1.

Answer 37

ok

Answer 38

nice graph by the way

Answer 39

With an amplitude of 2, it will instead go up to 2 and down to -2.