Some Trig Identities: Prove cos(4t) = 8cos^4(t) - 8cos^2(t) + 1, etc.

[Aminta_1900]

Hi,

I'm having trouble with a few of these trig identity problems. I'd appreciate it if someone could go through some of them step by step to give me a better understanding.

1. Prove the the following identity



The way I am going about this, where x here stands for theta, is starting with cos (2x + 2x) and applying the basic identities likes cos 2x*cos 2x - sin 2x*sin2x and cos^2 x - sin^x.....but I really can't see anything.

2. Prove the following identity



Again I have little idea, and expanding the double angle identities just makes it all look unwieldy, such as {1/(2 cos^2 x - 1)} + {(2 tan x)/(1 - tan^2 x)}.....

3. Simplify in terms of x



I have

tan^-1 x = a

tan a = x

sin(2 tan^-1 x) = sin(2a)

But then I can't seem to substitute tan a into any of the identities of sin 2a. I'm dabbling with tan a = (sin a)/(cos a) and sin 2a = 2 sin a cos a, but am getting nowhere.

4. Find the equation of the normal at pi/4 of the curve


Here I am getting the gradient correct but the y-intercept is wrong, and I'd just like to know if anyone can see why.

I have dy/dx = (3 cos x)/(-2 sin x) = (-3/2) cot x.

So (-3/2) cot 45deg = -3/2......so gradient of the normal = 2/3

To find the y coordinate where this normal crosses the curve I am taking the function as y = 3*sqr{ (4 - x^2)/4 } and the x value in radians as 0.785.

So y = 3*sqr{ (4 - 0.785^2)/4 } = 2.76

So I end up with y = (2/3)x + 2.24, but the correct answer is y = (2/3)x + { 5*sqr(2) }/6, i. e. c = 1.18.

Where have I gone wrong here. Any help will do, and thanks again for your time.

 

[khansaheb]

Hi,

I'm having trouble with a few of these trig identity problems. I'd appreciate it if someone could go through some of them step by step to give me a better understanding.

1. Prove the the following identity

View attachment 36395

The way I am going about this, where x here stands for theta, is starting with cos (2x + 2x) and applying the basic identities likes cos 2x*cos 2x - sin 2x*sin2x and cos^2 x - sin^x.....but I really can't see anything.

2. Prove the following identity

View attachment 36394

Again I have little idea, and expanding the double angle identities just makes it all look unwieldy, such as {1/(2 cos^2 x - 1)} + {(2 tan x)/(1 - tan^2 x)}.....

3. Simplify in terms of x

View attachment 36396

I have

tan^-1 x = a

tan a = x

sin(2 tan^-1 x) = sin(2a)

But then I can't seem to substitute tan a into any of the identities of sin 2a. I'm dabbling with tan a = (sin a)/(cos a) and sin 2a = 2 sin a cos a, but am getting nowhere.

4. Find the equation of the normal at pi/4 of the curve

View attachment 36397
Here I am getting the gradient correct but the y-intercept is wrong, and I'd just like to know if anyone can see why.

I have dy/dx = (3 cos x)/(-2 sin x) = (-3/2) cot x.

So (-3/2) cot 45deg = -3/2......so gradient of the normal = 2/3

To find the y coordinate where this normal crosses the curve I am taking the function as y = 3*sqr{ (4 - x^2)/4 } and the x value in radians as 0.785.

So y = 3*sqr{ (4 - 0.785^2)/4 } = 2.76

So I end up with y = (2/3)x + 2.24, but the correct answer is y = (2/3)x + { 5*sqr(2) }/6, i. e. c = 1.18.

Where have I gone wrong here. Any help will do, and thanks again for your time.

1. Prove the the following identity
View attachment 36395

Can you express cos(2Θ) in terms cos(Θ) and constants - only?

Please share your work.

2. Prove the following identity

Use

1 + sin(2Θ) = (sinΘ + cosΘ)²

continue......

 

[blamocur]

The way I am going about this, where x here stands for theta, is starting with cos (2x + 2x) and applying the basic identities likes cos 2x*cos 2x - sin 2x*sin2x and cos^2 x - sin^x.....but I really can't see anything.

Try [imath]\cos 2\theta = 2\cos^2\theta - 1[/imath].

 

[Dr.Peterson]

1. Prove the the following identity

The way I am going about this, where x here stands for theta, is starting with cos (2x + 2x) and applying the basic identities likes cos 2x*cos 2x - sin 2x*sin2x and cos^2 x - sin^x.....but I really can't see anything.

I'd just apply the double-angle identity for cosine (as in #3) twice.

2. Prove the following identity

Again I have little idea, and expanding the double angle identities just makes it all look unwieldy, such as {1/(2 cos^2 x - 1)} + {(2 tan x)/(1 - tan^2 x)}.....

One way is to start on the RHS, multiplying by the conjugate of the denominator. If you then think about double-angle identities, you can turn it into the LHS. If you start on the LHS, I'd rewrite it in terms of sine and cosine, then use [imath]\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)[/imath] for both denominators. Then simplify the resulting single fraction.

3. Simplify in terms of x

I have

tan^-1 x = a​

​

tan a = x​

​

sin(2 tan^-1 x) = sin(2a)​

But then I can't seem to substitute tan a into any of the identities of sin 2a. I'm dabbling with tan a = (sin a)/(cos a) and sin 2a = 2 sin a cos a, but am getting nowhere.

That's how I'd start; and I'd use [imath]\tan(a)=x[/imath] to find expressions for [imath]\sin(a)[/imath] and [imath]\cos(a)[/imath], to put into the expansion of [imath]\sin(2a)[/imath].

4. Find the equation of the normal at pi/4 of the curve

Here I am getting the gradient correct but the y-intercept is wrong, and I'd just like to know if anyone can see why.

I have dy/dx = (3 cos x)/(-2 sin x) = (-3/2) cot x.

So (-3/2) cot 45deg = -3/2......so gradient of the normal = 2/3

To find the y coordinate where this normal crosses the curve I am taking the function as y = 3*sqr{ (4 - x^2)/4 } and the x value in radians as 0.785.

So y = 3*sqr{ (4 - 0.785^2)/4 } = 2.76

So I end up with y = (2/3)x + 2.24, but the correct answer is y = (2/3)x + { 5*sqr(2) }/6, i. e. c = 1.18.

What does the intersection with the curve have to do with the y-intercept of the normal line? I have no idea what you are doing in that last part.

Just write the line using point-slope form and simplify!

 

[khansaheb]

1. Prove the the following identity

Use:

cos(2α) = 2 * cos²(α) - 1 .......... twice with proper substitution.

 

[Aminta_1900]

I'd just apply the double-angle identity for cosine (as in #3) twice.


One way is to start on the RHS, multiplying by the conjugate of the denominator. If you then think about double-angle identities, you can turn it into the LHS. If you start on the LHS, I'd rewrite it in terms of sine and cosine, then use [imath]\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)[/imath] for both denominators. Then simplify the resulting single fraction.


That's how I'd start; and I'd use [imath]\tan(a)=x[/imath] to find expressions for [imath]\sin(a)[/imath] and [imath]\cos(a)[/imath], to put into the expansion of [imath]\sin(2a)[/imath].


What does the intersection with the curve have to do with the y-intercept of the normal line? I have no idea what you are doing in that last part.

Just write the line using point-slope form and simplify!

Thanks a lot, I understand questions 1 and 3 now, but for question 2 I still need a little help in simplifying the fraction, as per the explanation you gave. I have it now as

 

[Aminta_1900]

Correction

 

[Harry_the_cat]

Correction

View attachment 36403

Note that the RHS only involves \(\displaystyle \theta\) not \(\displaystyle 2\theta\). That should tell you what to do with the \(\displaystyle \sin{2\theta}\).

Note that the denominator can be factorised as the difference of two squares. One of the factors is what you want on the denominator.

Can you see the next step?

 

[Aminta_1900]

I tried this before, but I can't seem to cancel anything out...

 

[Steven G]

Correction

View attachment 36403

Something is screaming out to me! One of the two factors in the left side denominators is in the denominator on the right side.
I would IMMEDIATELY multiply the right side so it would have that missing factor. Then you at least have the same denominators and now have to work on getting the numerators equal.

Personally I NEVER give up an opportunity to get both sides to look closer to one another.

Since (numerator of) the right hand side only has θ, now I would try to get all angles (in the numerator) to be θ or 2θ.

Everything I posted has already been said in other post, but I thought it would be nice for you to have it all in one post.

 

[Harry_the_cat]

You could do what Steven G says and start working with the RHS.

I would prefer to continue working with the LHS and factorise the diff of 2 squares on the denominator. Can you do that?
The following step is a bit tricky but good to have in your toolbox. Have a think about what you can do with the numerator?

 

[khansaheb]

Thanks a lot, I understand questions 1 and 3 now, but for question 2 I still need a little help in simplifying the fraction, as per the explanation you gave. I have it now as

View attachment 36401

sin(2x) + 1 = 2 * sin(x) * cos(x) + sin²(x) + cos²(x) = [sin(x) + cos(x)]²

continue....

 

[Aminta_1900]

Ok, thanks a lot fellas. I have it now.


It's just I've never been obliged to notice a difference of two squares in trigonometry before, nor would have thought to expand the 1 into sin^2 x + cos^2 x.

But then I hope my next query doesn't sound naive, but how did you get the (sin x + cos x)^2 from the 2 * sin x * cos x + sin^2 x + cos^2 x?

 

[Steven G]

2ab+a²+b² = (a+b)²
So 2sinxcos x + sin² x + cos² x = (sin x + cos x)²

 

[lookagain]

I'm having trouble with a few of these trig identity problems.

For any possible future problems, please post one problem per thread for
you and us to concentrate upon.