Trig equation problem

[jonnburton]

Hi I have a trig equation which I can't work out how to approach. I was wondering whether anyone could tell me how to do this.


This is the question:

Given that x is an acute angle and \(\displaystyle sin x = \frac {12}{13}\), use the identity \(\displaystyle cos^2x + sin^2 x = 1\) to show that \(\displaystyle 13 cos x= 5\)



I understand the identity given, but I don't know how to apply this to the equation given, which contains no squares.


I don't think you can eliminate the squares in the identity by taking the square root of all terms, so that \(\displaystyle \sqrt cos^2x + \sqrt sin^2 x = \sqrt 1\), which would equal \(\displaystyle cos x + sin x = 1\).

Having said that, I did try this, and on the off-chance that there was any sense to this I will post my working below:

\(\displaystyle sin x = 1 - cos x\)

Substituting this value in place of sin x in the original equation:

\(\displaystyle 1 - cos x = \frac {12}{13}\)

\(\displaystyle - cos x = -\frac {1}{13}\)

\(\displaystyle 13 cos x = 1\), a result which has not answered the question.

Many thanks in advance for any help.

 

[Deleted member 4993]

Hi I have a trig equation which I can't work out how to approach. I was wondering whether anyone could tell me how to do this.


This is the question:

Given that x is an acute angle and \(\displaystyle sin x = \frac {12}{13}\), use the identity \(\displaystyle cos^2x + sin^2 x = 1\) to show that \(\displaystyle 13 cos x= 5\)



I understand the identity given, but I don't know how to apply this to the equation given, which contains no squares.


For some reason (I can't see why not) I don't think you can eliminate the squares in the identity by taking the square root of all terms, so that \(\displaystyle \sqrt cos^2x + \sqrt sin^2 x = \sqrt 1\), which would equal \(\displaystyle cos x + sin x = 1\).

Having said that, I did try this, and on the off-chance that there was any sense to this I will post my working below:

\(\displaystyle sin x = 1 - cos x\) ← How is that?

it should be

sin²(x) = 1 - cos²(x)

Substituting this value in place of sin x in the original equation:

\(\displaystyle 1 - cos x = \frac {12}{13}\)

\(\displaystyle - cos x = -\frac {1}{13}\)

\(\displaystyle 13 cos x = 1\), a result which has not answered the question.

Many thanks in advance for any help.

.

 

[jonnburton]

OK I see that this bit is not correct: \(\displaystyle sin x = 1 - cos x\)

and that \(\displaystyle sin^2 (x)= 1 - cos^2 (x)\)

But I can't see how to work from this back to the original equation: \(\displaystyle sin x = \frac {12}{13}\)

In particular, I can't see how to get from \(\displaystyle 1 - cos^2 (x)\) to \(\displaystyle sin x\)

 

[Deleted member 4993]

OK I see that this bit is not correct: \(\displaystyle sin x = 1 - cos x\)

and that \(\displaystyle sin^2 (x)= 1 - cos^2 (x)\)

But I can't see how to work from this back to the original equation: \(\displaystyle sin x = \frac {12}{13}\)

In particular, I can't see how to get from \(\displaystyle 1 - cos^2 (x)\) to \(\displaystyle sin x\)

Really!!

1- cos²(x) = sin²(x) → √[sin²(x)] = sin(x)

 

[srmichael]

Hi I have a trig equation which I can't work out how to approach. I was wondering whether anyone could tell me how to do this.


This is the question:

Given that x is an acute angle and \(\displaystyle sin x = \frac {12}{13}\), use the identity \(\displaystyle cos^2x + sin^2 x = 1\) to show that \(\displaystyle 13 cos x= 5\)



I understand the identity given, but I don't know how to apply this to the equation given, which contains no squares.


I don't think you can eliminate the squares in the identity by taking the square root of all terms, so that \(\displaystyle \sqrt cos^2x + \sqrt sin^2 x = \sqrt 1\), which would equal \(\displaystyle cos x + sin x = 1\).

Having said that, I did try this, and on the off-chance that there was any sense to this I will post my working below:

\(\displaystyle sin x = 1 - cos x\)

Substituting this value in place of sin x in the original equation:

\(\displaystyle 1 - cos x = \frac {12}{13}\)

\(\displaystyle - cos x = -\frac {1}{13}\)

\(\displaystyle 13 cos x = 1\), a result which has not answered the question.

Many thanks in advance for any help.

Jonnburton:

\(\displaystyle \cos^2x+\sin^2x=1\)

\(\displaystyle \cos^2x+(\frac{12}{13})^2=1\)

\(\displaystyle \cos^2x+\frac{144}{169}=1\)

and continue...

 

[JeffM]

OK I see that this bit is not correct: \(\displaystyle sin x = 1 - cos x\)

and that \(\displaystyle sin^2 (x)= 1 - cos^2 (x)\)

But I can't see how to work from this back to the original equation: \(\displaystyle sin x = \frac {12}{13}\)

In particular, I can't see how to get from \(\displaystyle 1 - cos^2 (x)\) to \(\displaystyle sin x\)

You have been given two excellent ways to work this problem, but I suspect that it may be helpful to discuss what I suspect to be the source of your confusion.

When people start algebra, they have great difficulty realizing that the letters represent plain, ordinary, everyday numbers. Beginning students get into conceptual difficulty whenever they forget that basic fact.

But functions also just represent plain, ordinary, everyday numbers.

If I told you x = 3 and gave you a problem in which x squared appeared, you would see that x squared is 9 immediately. But sin(x) is just a number. In this case, you are told that sin(x) = 12/13. It's just a number. You know how to square numbers so you can easily figure out what the square of sin(x) is. Every time that you forget that you are just dealing with numbers, things will become confusing.

 

[jonnburton]

Many thanks for all of your replies and advice.

I can identify exactly with what JeffM has written: I do get distracted by algebraic terms, and would do well to remind myself that they just represent numbers. I hope that in the not-too-distant future I can 'see' these things as they are, and not sweat so much over what are very simple problems!

That said, I think I understand both explanations given by Subhotosh and srmichael. Both led me to the same conclusion, namely \(\displaystyle 13cos x = 1\) which made me think, maybe a bit hopefully, that perhaps there is a typing error in the book I am using.

The following is my following up of Subhotosh's help:

\(\displaystyle 1 -cos^2x = sin^2 x\)

\(\displaystyle 1 -cos x = sin x\)

\(\displaystyle 1 -cos x = \frac{12}{13}\)

\(\displaystyle -cos x = \frac{12}{13} - 1\)

\(\displaystyle -cos x = - \frac{1}{13}\)

\(\displaystyle cos x = \frac{1}{13}\)

\(\displaystyle 13cos x = 1\)


And, following up on srmichael's suggestion:


\(\displaystyle cos^2x + \frac{12}{13}^2 = 1 \)

\(\displaystyle cos^2x + \frac{144}{169} = 1 \)


\(\displaystyle cos x + \frac{12}{13} = 1 \)

\(\displaystyle cos x = 1- \frac{12}{13} \)

\(\displaystyle cos x = \frac{1}{13} \)

Which again leads to the same solution, that \(\displaystyle 13cos x = 1\), rather than the book's solution which is: \(\displaystyle 13cos x = 5\)

 

[srmichael]

Many thanks for all of your replies and advice.

I can identify exactly with what JeffM has written: I do get distracted by algebraic terms, and would do well to remind myself that they just represent numbers. I hope that in the not-too-distant future I can 'see' these things as they are, and not sweat so much over what are very simple problems!

That said, I think I understand both explanations given by Subhotosh and srmichael. Both led me to the same conclusion, namely \(\displaystyle 13cos x = 1\) which made me think, maybe a bit hopefully, that perhaps there is a typing error in the book I am using.

The following is my following up of Subhotosh's help:

\(\displaystyle 1 -cos^2x = sin^2 x\)

\(\displaystyle 1 -cos x = sin x\)

\(\displaystyle 1 -cos x = \frac{12}{13}\)

\(\displaystyle -cos x = \frac{12}{13} - 1\)

\(\displaystyle -cos x = - \frac{1}{13}\)

\(\displaystyle cos x = \frac{1}{13}\)

\(\displaystyle 13cos x = 1\)


And, following up on srmichael's suggestion:


\(\displaystyle cos^2x + \frac{12}{13}^2 = 1 \)

\(\displaystyle cos^2x + \frac{144}{169} = 1 \)


\(\displaystyle cos x + \frac{12}{13} = 1 \) INCORRECT!

\(\displaystyle cos x = 1- \frac{12}{13} \)

\(\displaystyle cos x = \frac{1}{13} \)

Which again leads to the same solution, that \(\displaystyle 13cos x = 1\), rather than the book's solution which is: \(\displaystyle 13cos x = 5\)

You cannot take the square root of individual terms when they are being added or subtracted from one another!! As Denis would say, Tatoo that on your wrist.

\(\displaystyle \cos^2x+\frac{144}{169}=1\)

\(\displaystyle \cos^2x=1-\frac{144}{169}\)

\(\displaystyle \cos^2x=\frac{169}{169}-\frac{144}{169}\)

\(\displaystyle \cos^2x=\frac{25}{169}\)

NOW you can take the square root of both sides:

\(\displaystyle cosx=\frac{5}{13}\)

\(\displaystyle 13cosx=5\)

 

[jonnburton]

Ah, OK I'll remember that srmichael! Thanks a lot!