Question

Myininaya, Satellite - Please help me solve these two Trigonometric Identities! Thanks!
cos4(x) = cos^4(x) - 6sin^2(x) cos^2(x) + sin^4(x)

sec^2(x) = 2sec(x) / sec(x) + 1

Answer 1

i am confused because it looks like you have
\[\cos^4(x)\]on both sides

Answer 2

the second cos^4x, the 4 is a power

Answer 3

like, cos to the fourth of x

Answer 4

ooh and is the first one
\[\cos(4x)\]?

Answer 5

Yes!! You got it.

Answer 6

I'm sorry

Answer 7

ok then i think it will work out fine if we use the "double angle formula" twice.
\[\cos(2x)=\sin^2(x)-\cos^2(x)\] among other things, so maybe if we try
\[\cos(4x)=\sin^2(2x)-\cos^2(2x)\] as a start and then use it again we will get what we want. let me try to write it on paper first before i say something incorrect

Answer 8

Yes, good idea. I see exactly how you figured that out. However, I am completely stuck after that.

Answer 9

SO we are working with the left side ?

Answer 10

yes this is what i have on paper so far
\[\cos(4x)=\cos^2(2x)-\sin^2(2x)\] and now the first term is
\[\cos^2(2x)=(\cos^2(x)-\sin^2(x))^2=\cos^4(x)-2\sin^2(x)\cos^2(x)+\sin^4(x)\]

Answer 11

now lets see what happens with the
\[\sin^2(2x)\]

Answer 12

yeah it works out just as you want it to

Answer 13

\[\sin(2x)=2\sin(x)\cos(x)\] and therefore
\[\sin^2(2x)=(2\sin(x)\cos(x))^2=4\sin^2(x)\cos^2(x)\]

Answer 14

subtract that off from the thing i wrote above and you get exactly what you want

Answer 15

you are truly my savior, thank you! Now, do you think you will be able to also help me with the second problem?

Answer 16

sure
did you get the last one because i wrote it out in stages? it is clear right?

Answer 17

Very clear. I got it.

Answer 18

second one is
\[\sec^2(x)=\frac{2\sec(x)}{\sec(x)+1}\]?

Answer 19

\[\sec ^{2 }u/2 = 2secu / secu + 1\]

Answer 20

Yes, you are right except for the left side should read sec squared times the quantity of X divided by 2.

Answer 21

I'm sorry, don't mind me using U's instead of X's.

Answer 22

oh it is
\[\sec^2(\frac{x}{2})\] on the left?

Answer 23

COrrect! Yes.

Answer 24

oh lord. ok now i have to think. but before we think, lets write out what the right hand side really is
\[\frac{\frac{2}{a}}{\frac{1}{a}+1}=\frac{2}{a+1}\] and so
\[\frac{2\sec(x)}{\sec(x)+1}=\frac{2}{\cos(x)+1}\]

Answer 25

Got it.

Answer 26

good because that actually means we are done!

Answer 27

That equals the left side?

Answer 28

\[\cos(\frac{x}{2})=\sqrt{\frac{1+\cos(x)}{2}}\]

Answer 29

so
\[\sec(\frac{x}{2})=\sqrt{\frac{2}{1+\cos(x)}}\]

Answer 30

and therefore
\[\sec^2(\frac{x}{2})=\frac{2}{1+\cos(x)}\] just as we hoped

Answer 31

you are awesome. This was so helpful. Thank you so, so much.

Answer 32

yw

Answer 33

Is there a way I could find you if I needed help in the future? Luckily I only have one more month of this semester until Trig is finished.

Answer 34

Thanks again.