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# Need help with this trigonometric question involving r cos(2x + a). Confused Watch

1. Solve the equation 2cosx + 5sinx= 4 by expressing in the form r cos(2x+a)
* (x is actually theta and a is actually alpha)

I don't know how to start this question. Would definitely be grateful if someone could help me Thanks
2. (Original post by bryan966)
Solve the equation 2cosx + 5sinx= 4 by expressing in the form r cos(2x+a)
* (x is actually theta and a is actually alpha)

I don't know how to start this question. Would definitely be grateful if someone could help me Thanks
Do you mean rcos(2x + a) or rcos(x + a) ?
3. (Original post by davros)
Do you mean rcos(2x + a) or rcos(x + a) ?
rcos(2x+a), if it was the latter I would probably have done it with no problem. So yeah
4. (Original post by bryan966)
rcos(2x+a), if it was the latter I would probably have done it with no problem. So yeah
Sorry, it's got to be a misprint unless I'm missing the blindingly obvious (always possible as my sleep has been disrupted by the storms for 2 nights running!!)
5. (Original post by davros)
Sorry, it's got to be a misprint unless I'm missing the blindingly obvious (always possible as my sleep has been disrupted by the storms for 2 nights running!!)

My sleep has also been disturbed (by bullet chess addiction) but I agree. Definitely a misprint.
6. Hmmm. Because originally, the question asked us to solve it by expressing it in the form of r cos(x+a). I did that, then the next question was the one I posted. Hahha, if it's a misprint then I'm burning the book... Thank you though
7. (Original post by bryan966)
Hmmm. Because originally, the question asked us to solve it by expressing it in the form of r cos(x+a). I did that, then the next question was the one I posted. Hahha, if it's a misprint then I'm burning the book... Thank you though
Any chance you can scan / photograph the page containing the two successive questions and upload?

No worries if you can't!
8. (Original post by davros)
Any chance you can scan / photograph the page containing the two successive questions and upload?

No worries if you can't!
Here you go
It's question number 5.
And thank you so very much
Attached Images

9. (Original post by bryan966)
Here you go
It's question number 5.
And thank you so very much
Hey, do you have the answers to it?
10. (Original post by Ollie_Brown)
Hey, do you have the answers to it?
Well, this is kinda funny... There is no answer
I'm pretty confused hhahaha. I assumed its the same answer from question number 4 which are 26'10', 110'14'. Sorry if this doesn't make sense.
But anyway, did you solve it?
11. (Original post by bryan966)
Well, this is kinda funny... There is no answer
I'm pretty confused hhahaha. I assumed its the same answer from question number 4 which are 26'10', 110'14'. Sorry if this doesn't make sense.
But anyway, did you solve it?
I'm a bit rusty. I got 110 and 26 from 4a) where do the others come from again? And I'm still having a go at it, I can't remember doing anything like that this year.
12. Well theta doesn't equal x. Theta's represents any angle.
That means you can solve it as usual letting 2x=theta and find values within double the range they give you: Rcos(2x+a)=Rcos(theta+a)
Then divide by 2 to find x. If that makes sense.
13. Yeah. I got that that too. It's number 5 that's making me wanna kill myself haha. Thank you
14. (Original post by cerlohee)
Well theta doesn't equal x. Theta's represents any angle.
That means you can solve it as usual letting 2x=theta and find values within double the range they give you: Rcos(2x+a)=Rcos(theta+a)
Then divide by 2 to find x. If that makes sense.
Hmmmm. Have you seen the attachment in one of the posts above that I posted? See question number 4 then 5. What I meant was rcos(2theta + a), I wrote x coz I didn't know how to type the theta symbol. I'm sorry if I'm the one who's confused.
15. (Original post by bryan966)
Here you go
It's question number 5.
And thank you so very much
I'm flummoxed I'm afraid - I can't see any way in which using 2x + a makes the equations easier, or even solvable!

The later question where it asks you to use cos(2x+a) makes more sense because you've got a cos^2 x term and a sinxcosx term so you can convert these to cos2x and sin2x and then rearrange.

I can only think question 5 is some sort of aberration on the part of the author unless your teacher has any ideas (or have you broken up now?).
16. How do you find r and a for the question you know how to do?

A cosx - B sinx = r cos(x + a)

Well, if we want to derive the formula, or show every single step, then you would start by expanding the cosine.

... = r cosx cosa - r sinx sina

Matching the coefficients of like terms gives A = r cosa and B = r sina

Then, by drawing the the right triangle with angle a and hypotenuse r, we find that a = arctan(B/A) and r = sqrt(A^2 + B^2).

Analogously, we probably want to solve this by expanding cos(2x + a). Then, judging by the instructions from problem 1...

Tell me if everything works out.
17. (Original post by davros)
I'm flummoxed I'm afraid - I can't see any way in which using 2x + a makes the equations easier, or even solvable!

The later question where it asks you to use cos(2x+a) makes more sense because you've got a cos^2 x term and a sinxcosx term so you can convert these to cos2x and sin2x and then rearrange.

I can only think question 5 is some sort of aberration on the part of the author unless your teacher has any ideas (or have you broken up now?).
Well, I won't be seeing my teacher until next week so asking him would take time. Thanks for helping me clear up on question 7 haha!
Back to q.5, I'm asking my mom's colleague to try and help me solve it
If she couldn't do it, then yeah maybe you're right. Thank you so so much
18. (Original post by aznkid66)
How do you find r and a for the question you know how to do?

A cosx - B sinx = r cos(x + a)

Well, if we want to derive the formula, or show every single step, then you would start by expanding the cosine.

... = r cosx cosa - r sinx sina

Matching the coefficients of like terms gives A = r cosa and B = r sina

Then, by drawing the the right triangle with angle a and hypotenuse r, we find that a = arctan(B/A) and r = sqrt(A^2 + B^2).

Analogously, we probably want to solve this by expanding cos(2x + a). Then, judging by the instructions from problem 1...

Tell me if everything works out.
Thanks, will try
19. (Original post by aznkid66)
How do you find r and a for the question you know how to do?

A cosx - B sinx = r cos(x + a)

Well, if we want to derive the formula, or show every single step, then you would start by expanding the cosine.

... = r cosx cosa - r sinx sina

Matching the coefficients of like terms gives A = r cosa and B = r sina

Then, by drawing the the right triangle with angle a and hypotenuse r, we find that a = arctan(B/A) and r = sqrt(A^2 + B^2).

Analogously, we probably want to solve this by expanding cos(2x + a). Then, judging by the instructions from problem 1...

Tell me if everything works out.
I think we can see that - but what are you going to equate the cos2x to or the sin2x? It doesn't seem obvious where the question is leading unless we're all missing something!
20. (Original post by davros)
I think we can see that - but what are you going to equate the cos2x to or the sin2x? It doesn't seem obvious where the question is leading unless we're all missing something!
Hey there. This is what my mom's colleague got. I'm a bit confused whether is a valid solution. A lil warning tho, it takes time to know which part comes first. Cheers
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