Trigonometric Identity Equations

[lual0209]

Hi. I'm struggling with a few questions - I seem to understand the trigonometric identities, but I'm not sure of the algebra. Can anyone please help solve the following equations (with steps and explanations, if possible):

1. VERIFY THE IDENTITY: sec^2 x sin^2 x - sec^2 x = -1

2. VERIFY THE IDENTITY: sin2x = 2tan x/1 + tan^2 x

3. SOLVE: 2sin^2 x-11cosx - 11 = 0, for 0 is less than or equal to x, which is less than 2 pi

4. SOLVE: 4sin x cos x - 2 square root 2sin x + 2cos x - square root 2 = 0, where x is a real number

5. VERIFY THE GIVEN IDENTITY: sin (x + y) sin (x - y) = sin^2 x - sin^2 y

Thank you!

 

[Deleted member 4993]

Hi. I'm struggling with a few questions - I seem to understand the trigonometric identities, but I'm not sure of the algebra. Can anyone please help solve the following equations (with steps and explanations, if possible):

1. VERIFY THE IDENTITY: sec^2 x sin^2 x - sec^2 x = -1 <<< Use sec(x) = 1/cos(x) and 1- sin[sup:37odpvmy]2[/sup:37odpvmy](x) = cos[sup:37odpvmy]2[/sup:37odpvmy](x)
2. VERIFY THE IDENTITY: sin2x = 2tan x/1 + tan^2 x <<< use 1 + tan[sup:37odpvmy]2[/sup:37odpvmy](x) = sec[sup:37odpvmy]2[/sup:37odpvmy](x)

3. SOLVE: 2sin^2 x-11cosx - 11 = 0, for 0 is less than or equal to x, which is less than 2 pi <<< use 1- sin[sup:37odpvmy]2[/sup:37odpvmy](x) = cos[sup:37odpvmy]2[/sup:37odpvmy](x) then use quadratic formula
4. SOLVE: 4sin x cos x - 2 square root 2sin x + 2cos x - square root 2 = 0, where x is a real number

4sin x cos x + 2cos x - 2 ?2 sin x - ?2 = 0

2cos(x) * [2sin(x) + 1] - ?2 * [2sin(x) + 1] = 0 .... now continue

5. VERIFY THE GIVEN IDENTITY: sin (x + y) sin (x - y) = sin^2 x - sin^2 y <<< use sin(x±y) = sin(x)*cos(y) ± cos(x)*sin(y)

Thank you!

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[lual0209]

Thanks Subhotosh.

Well, for Q. 1, I think that you have to factor, and you then end up with something like sec^2 x (1 - sin^2 x) or - sec^2 x (1 - sin^2 x). Now, 1 - sin^2 x = cos^2 x, so you would then have sec^2 x cos^2 x or - sec^2 x cos^2 x. Finally, as sec^2 x = 1/cos^2 x, it becomes 1/cos^2 x or - 1/cos^2 x * cos^2 x, and the cos^2x cancel, leaving the answer.

For Q. 2, I know that sin 2 x = 2 sin x cos x, and then I think you multiply by cos x/cos x, but I'm not sure.

For Q. 3, I get 2 (1 - cos^2 x) - 11 cos x - 11 = 0, then 2 - 2 cos^2 x - 11 cos x - 11 = 0, then 2 cos^2 x - 11 cos x -9 = 0 (I'm not sure how that works though). Beyong that, I'm not sure.

For Q. 4, I really don't know.

For Q. 5, I understand that sin (x + y) sin (x - y) = (sin x cos y + cos x sin y) (sin x cos y - cos x sin y), but then I'm not sure.

Thanks again for your help, Subhotosh.

 

[Deleted member 4993]

Thanks Subhotosh.

Well, for Q. 1, I think that you have to factor, and you then end up with something like sec^2 x (1 - sin^2 x) or - sec^2 x (1 - sin^2 x). Now, 1 - sin^2 x = cos^2 x, so you would then have sec^2 x cos^2 x or - sec^2 x cos^2 x. Finally, as sec^2 x = 1/cos^2 x, it becomes 1/cos^2 x or - 1/cos^2 x * cos^2 x, and the cos^2x cancel, leaving the answer.

For Q. 2, I know that sin 2 x = 2 sin x cos x, and then I think you multiply by cos x/cos x, but I'm not sure.

sin2x = 2tan x/(1 + tan^2 x)

RHS

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

= 2tan(x)/sec[sup:3qq51bos]2[/sup:3qq51bos](x)

= 2 * sin(x)/cos(x) * cos[sup:3qq51bos]2[/sup:3qq51bos](x)

= 2 * sin(x) * cos(x) = sin(2x)


For Q. 3, I get 2 (1 - cos^2 x) - 11 cos x - 11 = 0, then 2 - 2 cos^2 x - 11 cos x - 11 = 0, then 2 cos^2 x - 11 cos x -9 = 0 (I'm not sure how that works though). Beyong that, I'm not sure.

Above is a quadratic equation - use quadratic formula

Ax2 + Bx + C = 0

then

x[sub:3qq51bos]1,2[/sub:3qq51bos] = [-B ± ?(B[sup:3qq51bos]2[/sup:3qq51bos]-4AC)]/(2A)

If you don't remember this, dust-off your algebra book - study it and tattoo it in your brain - you'll use it over and over.

For Q. 4, I really don't know.

Factorize it and solve it - simple algebra.

For Q. 5, I understand that sin (x + y) sin (x - y) = (sin x cos y + cos x sin y) (sin x cos y - cos x sin y),

Do you see the form (a+b)(a-b)?

Then use

(a+b)(a-b) = a[sup:3qq51bos]2[/sup:3qq51bos] - b[sup:3qq51bos]2[/sup:3qq51bos]

And continue....but then I'm not sure.

Thanks again for your help, Subhotosh.

 

[lual0209]

I think I've managed to work all the questions out except Q. 4. I do know that sin alpha cos beta = 1/2 [sin (alpha + beta) + sin (alpha - beta)], but I'm not sure how to apply that to this particular equation, or how to work out the rest of it.

Here it is again for clarity:

4. SOLVE: 4sin x cos x - 2?2sin x + 2cos x - ?2 = 0, where x is a real number

Thank you.

 

[fasteddie65]

2 sin^2 x - 11 cos x - 11 = 0
2 (1 - cos^2 x) - 11 cos x - 11 = 0
2 - 2 cos^2 x - 11 cos x - 11 = 0
-2 cos^2 x - 11 cos x - 9 = 0
2 cos^2 x + 11x + 9 = 0
We are looking for factors of 18 (2 • 9) whose sum is 11: 2 and 9
Rewrite as four terms: 2 cos^2 x + 2 cos x + 9 cos x + 9 = 0
2 cos x (cos x + 1) + 9 (cos x + 1) = 0
(cos x + 1)(2 cos x + 9) = 0
cos x = -1 or cos x = -9/2
The first equation has one solution in [0, 2?), namely x = ?.
The second equation has no solutions.

Why use Quadratic Formula when factoring is not that bad?

 

[Deleted member 4993]

I think I've managed to work all the questions out except Q. 4. I do know that sin alpha cos beta = 1/2 [sin (alpha + beta) + sin (alpha - beta)], but I'm not sure how to apply that (not needed here) to this particular equation, or how to work out the rest of it.

Here it is again for clarity:

4. SOLVE: 4sin x cos x - 2?2sin x + 2cos x - ?2 = 0, where x is a real number

4sin x cos x + 2cos x - 2 ?2 sin x - ?2 = 0

2cos(x) * [2sin(x) + 1] - ?2 * [2sin(x) + 1] = 0

[2sin(x) + 1] [sub:le9sle4i]*[/sub:le9sle4i] [2cos(x) - ?2] = 0

continue.....




Thank you.

 

[lual0209]

I think I've managed to work all the questions out except Q. 4. I do know that sin alpha cos beta = 1/2 [sin (alpha + beta) + sin (alpha - beta)], but I'm not sure how to apply that (not needed here) to this particular equation, or how to work out the rest of it.

Here it is again for clarity:

4. SOLVE: 4sin x cos x - 2?2sin x + 2cos x - ?2 = 0, where x is a real number

4sin x cos x + 2cos x - 2 ?2 sin x - ?2 = 0

2cos(x) * [2sin(x) + 1] - ?2 * [2sin(x) + 1] = 0

[2sin(x) + 1] [sub:1ga9h9w8]*[/sub:1ga9h9w8] [2cos(x) - ?2] = 0

continue.....

2 sin x + 1 = 0, 2 sin x = -1, 2 sin x/2 = -1/2, sin x = -1/2 (solutions are 7pi/6, 210 deg, 11pi/6, 330 deg).
2 cos x - ?2 = 0, 2 cos x = ?2, 2 cos x/2 = ?2/2, cos x = ?2/2 (solutions are pi/4, 45 deg, 7pi/4, 315 deg).
Am I right? Thank you.


Thank you.

 

[Deleted member 4993]

I think I've managed to work all the questions out except Q. 4. I do know that sin alpha cos beta = 1/2 [sin (alpha + beta) + sin (alpha - beta)], but I'm not sure how to apply that (not needed here) to this particular equation, or how to work out the rest of it.

Here it is again for clarity:

4. SOLVE: 4sin x cos x - 2?2sin x + 2cos x - ?2 = 0, where x is a real number

4sin x cos x + 2cos x - 2 ?2 sin x - ?2 = 0

2cos(x) * [2sin(x) + 1] - ?2 * [2sin(x) + 1] = 0

[2sin(x) + 1] [sub:3p4uz5nk]*[/sub:3p4uz5nk] [2cos(x) - ?2] = 0

continue.....

2 sin x + 1 = 0, 2 sin x = -1, 2 sin x/2 = -1/2, sin x = -1/2

Very good.

Since your domain was all real 'x' then your answer is:

x = 7?/6 ± 2n? ( where n = 0,1,2....)

x = 11?/6 ± 2n? (where n = 0,1,2....)


(solutions are 7pi/6, 210 deg, 11pi/6, 330 deg).
2 cos x - ?2 = 0, 2 cos x = ?2, 2 cos x/2 = ?2/2, cos x = ?2/2

x = ± ?/4 ± 2n? ( where n = 0,1,2....)


(solutions are pi/4, 45 deg, 7pi/4, 315 deg).
Am I right? Thank you.


Thank you.