if 7sin^2theta + 3cos^2 theta = 4 prove that sec theta + cos theta = 2 +2/3

Guest Oct 5, 2013

#1**0 **

1. 7sin^2theta + 3cos^2 theta = 4 => 7sin^2theta + 3cos^2 theta = 4(sin^2 theta + cos^2 theta) => 3sin^2 theta - cos^2 theta = 0.Arrived at homogeneous trigonometric equation, and since sin^2 theta is not equal to cos^2 theta and both are not equal to 0, we can divide the components by cos^2 theta:

3tg^2 theta = 1 =>tg theta = +-1/sqrt(3). Theta = +-arctg(1/sqrt(3)) + pi n = +- 30 degrees (or +-pi/6) + pi n

2. cosine is a symmetric function, therefore cos(-pi/6) = cos pi/6, hence sec theta + cos theta = 1/cos(pi/6) + cos (pi/6) = 2/sqrt(3) + sqrt(3)/2 = (4 + 3)/2sqrt(3) = 7/sqrt(3)

Your presentation of the task contains either methodic or typewriting error since 2 + 2/3 = 8/3 can be written as a single number, but in the context of this task this looks illogically.

3tg^2 theta = 1 =>tg theta = +-1/sqrt(3). Theta = +-arctg(1/sqrt(3)) + pi n = +- 30 degrees (or +-pi/6) + pi n

2. cosine is a symmetric function, therefore cos(-pi/6) = cos pi/6, hence sec theta + cos theta = 1/cos(pi/6) + cos (pi/6) = 2/sqrt(3) + sqrt(3)/2 = (4 + 3)/2sqrt(3) = 7/sqrt(3)

Your presentation of the task contains either methodic or typewriting error since 2 + 2/3 = 8/3 can be written as a single number, but in the context of this task this looks illogically.

scrutinizer Oct 5, 2013

#2**0 **

Hi scrutinizer,

It is no big deal but you missed a 2 in your final answer

(4 + 3)/2sqrt(3) = 7/ 2sqrt(3)

I also agree that 2+2/3 is an input error.

I also think that you have missed out half the solution.

I did it a bit differently

I said

7sin^2theta + 3cos^2 theta = 4

4sin^2theta + 3sin^2theta + 3cos^2 theta = 4

4sin^2theta + 3(sin^2theta + cos^2 theta) = 4

4sin^2theta + 3*1 = 4

4sin^2theta = 1

sin^2theta = 1/4

sin theta = +-1/2

theta = 30,150, 210, 330, ....degrees

cos theta = +-(root3) / 2

so sectheta+costheta = +-( 2/root3 + root3/2) = +- [7/(2root3) ] = +- ( 7root3/6 )

If you think I am wrong I would be happy to listen to your argument.

It is no big deal but you missed a 2 in your final answer

(4 + 3)/2sqrt(3) = 7/ 2sqrt(3)

I also agree that 2+2/3 is an input error.

I also think that you have missed out half the solution.

I did it a bit differently

I said

7sin^2theta + 3cos^2 theta = 4

4sin^2theta + 3sin^2theta + 3cos^2 theta = 4

4sin^2theta + 3(sin^2theta + cos^2 theta) = 4

4sin^2theta + 3*1 = 4

4sin^2theta = 1

sin^2theta = 1/4

sin theta = +-1/2

theta = 30,150, 210, 330, ....degrees

cos theta = +-(root3) / 2

so sectheta+costheta = +-( 2/root3 + root3/2) = +- [7/(2root3) ] = +- ( 7root3/6 )

If you think I am wrong I would be happy to listen to your argument.

Melody Oct 5, 2013

#3**0 **

These two solutions are absolutely equivalent. The only difference was that I had passed to a homogeneous equation, while you preferred to transform it algebraicly. Just matter of your taste to choose the way of solution. Also, the final result of my calculation is identical to yours. I just didn't pay much significance to rationalizing that fraction.

scrutinizer Oct 17, 2013

#4**0 **

You are correct to say that the 2 solutions are almost equivalent.

However, when I said that you omited half the answer I was referring to the fact that you found only 7/[2sqrt(3)] and failed to identify -7/[2sqrt(3)] as a solution.

I also think that you misused the word symmetrical, I think you meant that cosine is an even function. Which is certainly true. You have answered correctly for theta = +- (pi/6) +2pi* k (where k is an integer)

but you have ignored the answers when theta = (pi +- pi/6 ) + 2pi* k (where k is an integer)

For example If theta = 5pi/6

7sin^{2}theta + 3cos ^{2}theta = 4 (first statement true)

sec theta + cos theta = -2/root3 - root3/2 = -7/(2root3)

So I repeat

sec theta+cos theta = +-( 2/root3 + root3/2) = +- [7/(2root3) ]

However, when I said that you omited half the answer I was referring to the fact that you found only 7/[2sqrt(3)] and failed to identify -7/[2sqrt(3)] as a solution.

I also think that you misused the word symmetrical, I think you meant that cosine is an even function. Which is certainly true. You have answered correctly for theta = +- (pi/6) +2pi* k (where k is an integer)

but you have ignored the answers when theta = (pi +- pi/6 ) + 2pi* k (where k is an integer)

For example If theta = 5pi/6

7sin

sec theta + cos theta = -2/root3 - root3/2 = -7/(2root3)

So I repeat

sec theta+cos theta = +-( 2/root3 + root3/2) = +- [7/(2root3) ]

Melody Oct 18, 2013

#5**0 **

To be honest, I trusted an online Russian-English translator that suggested both terms of English analogues of that notion, specifying that the term "symmetric" is applied most commonly to the notion of function and that indeed seemed rather substantiated as it directly corresponds to the graphic interpretation as cosine graph y = cos x is symmetric in respect of 0y axis while the word "even" - as I was able to draw- is used mostly in conjunction with numbers.

"you have ignored the answers when theta = (pi +- pi/6 ) +- 2pi" - this is totally unclear to me and seems quite "overloaded". How did the expression (pi +- pi/6 ) +- 2pi appear? The common formula for the roots of tangential equation is x = arctan y + pi n, where pi is a minimal positive period of tan. Further on, as cosine is symmetric or, if you prefer more, "even" then whether the argument positive or negative, the cosine is always positive. The value of theta was found via transformation to homogeneous via transformation in turn to tangential equation. It had two solutions - positive and negative. sec theta = 1/cos theta. As it follows from above, cos theta is always positive. Hence, the final result in this case is also positive. You preferred to find the roots via sinusoidal equation. So to get cos theta you as I suppose applied the main trigonometric identity, where it's not important if sine is positive or negative. You may assert that the root has two signes, but there's a question: is there indication of what a quadrant of Cartesian coordinate system should be considered? No, and it means both values must be considered and thus I logically came up with denial of myself You were right, but how it happened that I missed that solution? Mine looks formally correct.

"you have ignored the answers when theta = (pi +- pi/6 ) +- 2pi" - this is totally unclear to me and seems quite "overloaded". How did the expression (pi +- pi/6 ) +- 2pi appear? The common formula for the roots of tangential equation is x = arctan y + pi n, where pi is a minimal positive period of tan. Further on, as cosine is symmetric or, if you prefer more, "even" then whether the argument positive or negative, the cosine is always positive. The value of theta was found via transformation to homogeneous via transformation in turn to tangential equation. It had two solutions - positive and negative. sec theta = 1/cos theta. As it follows from above, cos theta is always positive. Hence, the final result in this case is also positive. You preferred to find the roots via sinusoidal equation. So to get cos theta you as I suppose applied the main trigonometric identity, where it's not important if sine is positive or negative. You may assert that the root has two signes, but there's a question: is there indication of what a quadrant of Cartesian coordinate system should be considered? No, and it means both values must be considered and thus I logically came up with denial of myself You were right, but how it happened that I missed that solution? Mine looks formally correct.

scrutinizer Oct 18, 2013

#6**0 **

Firstly my last entry had an omission in it that added to your confusion.

I have now added it in red. I have also changed +-2pi*k to just +2pi*k because since k can be a positive or a negative integer the - was not necessary.

Sorry about the confusion.

You had

theta = +-pi/6 + pi*n

I had

theta = +-pi/6 + 2pi*k OR (pi+-pi/6) + 2pi*k

These are both the same. I just seperated them because they would give different answers for cos.

Your logic is fine. You simply didn't think about cos(5pi/6)=cos(-5pi/6)= -1/root3

Hence you missed the second solution.

Now for some English terms

Even numbers are 2,4,6,...

Odd numbers are 1,3,5,.....

An even function is a function which is symmetrical about the y axis. (The axis of symmetry is x=0)

Hence f(-x) = f(x)

For example y=cos x OR y=x^{2}

An odd function is a function where f(-x) = - f(x)

For example y=sin x OR y=x^{3}

Odd functions are said to have point symmetry because if you rotate the the function 180 degrees about the origin (0,0) then it sits back over itself. I hope that makes sense to you.

You taught me a new word too. I had to look up what homogeneous equations were. I don't know where the word homogeneous comes from, I guess it refers to the fact that they all end up the same ratio, that is tan.

I have now added it in red. I have also changed +-2pi*k to just +2pi*k because since k can be a positive or a negative integer the - was not necessary.

Sorry about the confusion.

You had

theta = +-pi/6 + pi*n

I had

theta = +-pi/6 + 2pi*k OR (pi+-pi/6) + 2pi*k

These are both the same. I just seperated them because they would give different answers for cos.

Your logic is fine. You simply didn't think about cos(5pi/6)=cos(-5pi/6)= -1/root3

Hence you missed the second solution.

Now for some English terms

Even numbers are 2,4,6,...

Odd numbers are 1,3,5,.....

An even function is a function which is symmetrical about the y axis. (The axis of symmetry is x=0)

Hence f(-x) = f(x)

For example y=cos x OR y=x

An odd function is a function where f(-x) = - f(x)

For example y=sin x OR y=x

Odd functions are said to have point symmetry because if you rotate the the function 180 degrees about the origin (0,0) then it sits back over itself. I hope that makes sense to you.

You taught me a new word too. I had to look up what homogeneous equations were. I don't know where the word homogeneous comes from, I guess it refers to the fact that they all end up the same ratio, that is tan.

Melody Oct 20, 2013

#7**0 **

I'm aware more than you might suppose of what is even/uneven speaking both of numbers and functions.

The reason for my use of the term I chosed is already explained above.

"You had

tan theta = +-pi/6 + pi*n

I had

tan theta = +-pi/6 + 2pi*k OR (pi+-pi/6) + 2pi*k"

As far as I remember it's you who had sin and me who had tan.

Once again, may consider me futile but I can't get into your notation. Whether you have pi/6 or -pi/6 you have to add the period of tan function which is pi, not 2 pi. At least that's what I'd been taught. I think the notation have to be simplier,that is according to the main rules of roots notation. As for 5pi/6 or -5pi/6 or 9pi/6 or -9pi/6, the part "+ pi n" speaks for itself: you get all these angles when consecutively add other multiples of these original angles. When you have cos theta, then you have to insert either pi/6 or -pi/6. Since cos is "even" cos (-pi/6)= cos pi/6.

Homogeneous expression is the polynomial of which every term's sum of indexes of power is the same, for example, a^2 + ab + b^2. As a rule the equations of such a type are solved by dividing every term by the variable of the major power.

The reason for my use of the term I chosed is already explained above.

"You had

tan theta = +-pi/6 + pi*n

I had

tan theta = +-pi/6 + 2pi*k OR (pi+-pi/6) + 2pi*k"

As far as I remember it's you who had sin and me who had tan.

Once again, may consider me futile but I can't get into your notation. Whether you have pi/6 or -pi/6 you have to add the period of tan function which is pi, not 2 pi. At least that's what I'd been taught. I think the notation have to be simplier,that is according to the main rules of roots notation. As for 5pi/6 or -5pi/6 or 9pi/6 or -9pi/6, the part "+ pi n" speaks for itself: you get all these angles when consecutively add other multiples of these original angles. When you have cos theta, then you have to insert either pi/6 or -pi/6. Since cos is "even" cos (-pi/6)= cos pi/6.

Homogeneous expression is the polynomial of which every term's sum of indexes of power is the same, for example, a^2 + ab + b^2. As a rule the equations of such a type are solved by dividing every term by the variable of the major power.

scrutinizer Oct 22, 2013