Question

Helpp

Answer 1

@zarkam21 were you able to rewrite the expression as a product of two squares ?

Answer 2

sorry but i just given an example and really i dont see clearly the power of sin on this posted image -

Answer 3

So the 4 in \(\sin^{4}x\). What do you suppose the 4 means in this case?

Answer 4

like 4 times sin

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
@zarkam21 for example

sin^4 x = sin^2 x *sin^2 x
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 6

So this is what you were asking?

Answer 7

Basically, \(\sin^{4}x = (\sin^{2}x)(\sin^{2}x)\)

Answer 8

So there you go. That's like a freebie for you.

Answer 9

okay so that would be the answer to power as a product of two squared terms

Answer 10

Yes, correct. The next step would be to find the identity that corresponds to \(\sin^2x\)

Answer 11

2 sin x cos x

Answer 12

That's the identity for \(\sin(2x)\) which is a double angle identity which is different from \(\sin^2x\)

Answer 13

1-cos^2 x

Answer 14

=)

Answer 15

Yes, much better

Answer 16

so thats it right for this step =) now on multiplying the terms

Answer 17

You'd have to insert that expression in place of the previous for both instances of \(\sin^2x\)

Answer 18

I have a question though. Where is the original equation associated with this trigonometric manipulation?

Answer 19

sin4x=(1-cos^2 x)(1-cos^2 x)

Answer 20

That's the original equation?

Answer 21

sin^4x

Answer 22

Are you sure about that? Post the image of the original equation.

Answer 23

Okay, I see now. So the step where you are currently is
\(\sin^4x= (1 - \cos^2x)(1 - \cos^2x)\) And they want you to multiply that out.

Answer 24

The RHS that is (Right Hand Side)

Answer 25

I would suggest that you use the distributive property. Substitution can also be useful in performing this step.

Answer 26

sin 4x=1-cos(2)(x)^2

Answer 27

What method did you use to multiply \((1 - \cos^2x)(1 - \cos^2x)\)?

Answer 28

Distributive Property, FOIL, Stack Multiplication, Substitution ...?

Answer 29

hold on. i made a mistake

Answer 30

multiplication

Answer 31

Okay, see if you can post your steps here. You can either use the drawing board or \(\LaTeX\)

Answer 32

Or you can write it on paper and post a screenshot.

Answer 33

So are you saying that you're finished multiplying?

Answer 34

sin^4(x)

Answer 35

Is English your 1st or 2nd language @zarkam21 I'm asking if you think you're done multiplying \((1 - \cos^2x)(1 - \cos^2x)\)

Answer 36

Yes

Answer 37

English is my first , fyi

Answer 38

Okay, so do you understand what it is I'm asking you? If so, paraphrase it.

Answer 39

Your asking me if i am done multiplying the two equations that we have here

Answer 40

And the answer to that is what?

Answer 41

I simplified it down and got sin^4(x)

Answer 42

is that not the answer?

Answer 43

Except that isn't the goal. The goal is to express the answer in terms of powers of cosine following the sequence of steps they have laid out for you.

Answer 44

ugh

Answer 45

i thought you told me to multiply so i multiplied and got that

Answer 46

Yes, but when you multiply two binomials, you have to use the distributive rule for that.

Answer 47

\(a(b + c) = ab + ac\)

Answer 48

Do you know how to multiply \((1 - x)(1 - x)\)

Answer 49

If so, show me the steps.

Answer 50

OKay..

Answer 51

(1-x)(1-x)

combine like terms

(1-x)(1-x)=(1-x)^2

Answer 52

Multiply implies expanding, not simplifying.

Answer 53

At least not in this instance.

Answer 54

okay i wil use FOIL THEN

Answer 55

LOOK:

Answer 56

YES, you use FOIL

Answer 57

(1-x)(1-x)
1-2x+x^2

Answer 58

Still wrong?

Answer 59

It's either FOIL or distributive property for multiplying two binomials. When they say multiply they mean EXPAND it. So try again with multiplying \((1 - \cos^2x)(1 - \cos^2x)\).

Answer 60

\[\sin^2(x)(1-\cos^2(x))\]

Answer 61

Did you apply FOIL?

Answer 62

\[\sin^2(x)\sin^2(x)\]

Answer 63

FOIL. No substitution. You already showed me you know how to FOIL.

Answer 64

\[sinx ^{2+2}\]

Answer 65

ugh

Answer 66

Im expanding it

Answer 67

Remember, the goal is to express in terms of COSINE and NOT sine.

Answer 68

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Hero
It's either FOIL or distributive property for multiplying two binomials. When they say multiply they mean EXPAND it. So try again with multiplying \((1 - \cos^2x)(1 - \cos^2x)\).
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 69

isn't that what you asked

Answer 70

Yes EXPAND using FOIL and Distributive Property, NOT substitution.

Answer 71

Try again.

Answer 72

using foil i got: \[(-\cos^2(x)+1)(-\cos^2(x)+1)=\cos^4(x)-2\cos^2(x)+1\]

Answer 73

Am I right now

Answer 74

Wow, you actually got it finally.

Answer 75

Still though you have to now apply the squared identity for cosine next.

Answer 76

dont know how to do that

Answer 77

By the way, in case you didn't know,
\(\cos^2x = \dfrac{1}{2}[\cos(2x)+1]\)

Answer 78

oh okay

Answer 79

Also, BTW, the expression you wrote can be re-written as:
\((\cos^2x)^2 - 2\cos^2x + 1\)

Answer 80

So now all you have to do is just replace \(\cos^2x\) with its corresponding identity.

Answer 81

1-sin^2x

Answer 82

I have a question for you:
What do you think it means when the problem says "Re-write \(\sin^4x\) so that it involves only the first power of cosine."?

Answer 83

that the problem is simplified to have only the first power of cos

Answer 84

So that means at every step we're moving away from any use of sine.

Answer 85

Or at least that's the goal.

Answer 86

correct because we have to end up with cos

Answer 87

Yes, so that means any use of sin going forward is probably taboo.

Answer 88

Well sort of.

Answer 89

There may be an instance where we would use it again, but for now, no.

Answer 90

We still have to re-write the expression \((\cos^2x)^2 - 2\cos^2x + 1\) given the following identity:

\(\cos^2x = \dfrac{1}{2}[\cos(2x)+1]\)

Answer 91

\[1/2[\cos(2x)+1 ] 2 −2 (\frac{ 1 }{ 2} [\cos(2x)+1]\]

Answer 92

Hang on...

Answer 93

ok

Answer 94

How much time do you have to submit this?

Answer 95

i guess we can do it tomorrow or something doesn't need to be done right this instant

Answer 96

Actually, I have it and it was much more involved than I anticipated.

Answer 97

Going to type all this out ....

Answer 98

\(\begin{align*}
\sin^4x &= (\sin^2x)(\sin^2x)
\\&=(1 - \cos^2x)(1 - \cos^2x)
\\&=1 - 2\cos^2x + \cos^{4}x
\\&=1 - 2\cos^2x + (\cos^2x)^2
\\&=1 - 2\left[\dfrac{1}{2}\left(\cos(2x)+1\right)\right]+\left(\dfrac{1}{2}\left(\cos(2x)+1\right)\right)^2
\end{align*}\)

Have this so far ....

Answer 99

Continuing from that last step
\(\begin{align*}\\&=1 - 2\left[\dfrac{1}{2}\left(\cos(2x)+1\right)\right]+\left(\dfrac{1}{2}\left(\cos(2x)+1\right)\right)^2
\\&=1 - \cos(2x) + 1+\left(\dfrac{\cos(2x)}{2}+\dfrac{1}{2}\right)^2
\\&=2 - \cos(2x)+ \left(\dfrac{\cos(2x)+1}{2}\right)^2
\\&=2-\cos(2x)+\left(\dfrac{(\cos(2x)+1)^2}{4}\right)
\\&=2 - \cos(2x)+\dfrac{1}{4}(\cos(2x) + 1)^2
\end{align*}\)

Have that so far ...

Answer 100

yeah im just studying what you re doing. so if im not replying you know why

Answer 101

Continuing from the last step ....
\(\begin{align*}
\\&=2 - \cos(2x)+\dfrac{1}{4}(\cos(2x) + 1)^2
\\&=2 - \cos(2x)+\dfrac{1}{4}(\cos^2(2x)+2\cos(2x)+1)
\\&=2 - \cos(2x)+\dfrac{1}{4}\cos^2(2x)+\dfrac{1}{2}\cos(2x)+\dfrac{1}{4}
\\&=2 - \cos(2x) + \dfrac{1}{4}\left(\dfrac{1}{2}\left[\cos(4x)+1\right]\right)+\dfrac{\cos(2x)}{2}+\dfrac{1}{4}
\end{align*}\)

Answer 102

You could simplify further from there, but no need. The expression is finally expressed in terms of just the first power of cosine.

Answer 103

Okay so lol this is only completion of the first power of cosine

Answer 104

Ha Joking

Answer 105

It was much more involved than I originally thought for sure.

Answer 106

Well def looks like it :/ I thought we would be done

Answer 107

Let me see if I can simplify it further so it matches the expression on wolfram alpha:
http://www.wolframalpha.com/input/?i=sin%5E4(x)

Answer 108

Continuing from above:
\(\begin{align*}
\\&=2 - \cos(2x) + \dfrac{1}{4}\left(\dfrac{1}{2}\left[\cos(4x)+1\right]\right)+\dfrac{\cos(2x)}{2}+\dfrac{1}{4}
\\&=2 +\dfrac{2}{4}+\dfrac{\cos(2x)}{2}-\dfrac{2\cos(2x)}{2}+\dfrac{1}{8}\cos(4x)
\\&=\dfrac{10}{4}-\dfrac{\cos(2x)}{2}+\dfrac{1}{8}\cos(4x)
\\&=\dfrac{1}{8}(20 - 4\cos(2x)+\cos(4x))
\\&=\dfrac{1}{8}(-4\cos(2x)+\cos(4x) + 20)
\end{align*}\)

Answer 109

I'm only off by about 17 units

Answer 110

wow, is this the simplest form

Answer 111

Yes