MHB Prove Trigonometric Identities

[mitaka90]

I got this problem on my term test and it's the first problem I couldn't solve on a test ever since I'm in High School. I've tried to solve it at home even, but I still couldn't manage. The thing is that it doesn't even look difficult, maybe there's something I'm not seeing, so I hope someone can help me. I apologize that I don't know how to use this Latex thing to write the problem, so I'll just post a link where you could see the identity from an online calc I used only to display it.

https://www.symbolab.com/solver/logarithmic-equation-calculator/8sin^{4}α=cos4α + 4cos2α +3

I'm trying to prove that the right side is equal to the left and I manage to come to a point where the identity looks like this:

https://www.symbolab.com/solver/log...culator/sin^{4}α=1-sin^{2}α.cos^{2}α-sin^{2}α (I got rid of the 8)

The problem requires Double-Angle and Half-Angle Formulas.

 

[Greg]

If the intended problem is

$$8\sin^4(a)=\cos4a+4\cos2a+3$$

then there is a mistake somewhere. The above is not an identity.

 

[mitaka90]

If the intended problem is

$$8\sin^4(a)=\cos4a+4\cos2a+3$$

then there is a mistake somewhere. The above is not an identity.

What? Really? I'm pretty sure that it was like this. I guess I'm going to have to wait until my math teacher gives the term test back and see what she has to say. Or I might have remembered the identity incorrectly. I'll update this thread on Monday hopefully if she returns the term test. Thanks, though!

P.S. How are you sure it's not an identity?

 

[Greg]

I ran it through a calculator and the two sides were not equal. You're off by a sign:

$$8\sin^4a=\cos4a-4\cos2a+3$$

 

[kaliprasad]

What? Really? I'm pretty sure that it was like this. I guess I'm going to have to wait until my math teacher gives the term test back and see what she has to say. Or I might have remembered the identity incorrectly. I'll update this thread on Monday hopefully if she returns the term test. Thanks, though!

P.S. How are you sure it's not an identity?

if you put $\alpha = 0$ LHS =0 and RHS=8 so they are not same.

I am sorry did not see that greg has replied.

 

[Opalg]

It should be

$$8\sin^4(a)=\cos(4a) - 4\cos(2a) + 3.$$

You can prove that using the identities $\cos(2a) = 2\cos^2a - 1 = 1 - 2\sin^2a$. Then $$\cos(4a) = 2\cos^2(2a) - 1 = 2(1 - 2\sin^2a)^2 - 1 = 8\sin^4a - 8\sin^2a + 1,$$ and it follows that $$\cos(4a) - 4\cos(2a) = 8\sin^4a - 8\sin^2a + 1 - 4(1 - 2\sin^2a) = 8\sin^4a - 3.$$

Edit. Didn't see Greg's post at #4!

 

[MarkFL]

Essentially the same, but going in the other direction:

Consider:

$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$

Then:

$$8\sin^4(a)=8\left(\frac{1-\cos(2a)}{2}\right)^2=2-4\cos(2a)+2\cos^2(2a)$$

Consider next:

$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

Then:

$$8\sin^4(a)=2-4\cos(2a)+1+\cos(4a)=\cos(4a)-4\cos(2a)+3$$

 

[mitaka90]

I see, I might've forgotten that there was a minus sign. Then everything comes as clear as a day. Whew. Ty all!

btw, What kind of calculator did you use greg?

 

[Greg]

It's called PARI/gp. It's a free CAS with a number theory focus. Here's a link to the home page.

 

[mitaka90]

It's called PARI/gp. It's a free CAS with a number theory focus. Here's a link to the home page.

I'll check it out.

It turned out that it was indeed a minus sign. I got some points from the beginning of the problem and I managed to get the A or Excellent mark. I don't really know why I messed it up, but whatevs. Job's done!