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Solving Cos2a = Sina

Hi there,

I am trying to solve the following two (separate) questions.

i) SinA= Cos2A

ii) Sin2A -1 = Cos2A

Where A must be presented in radians

Find A.

I'm afraid I'm a bit rusty on trigonometry as I have not done this in the 3 years I have been at Uni, so couldn't find many helpful resources on how to get started on this problem.
Original post by VincentCheung
Hi there,

I am trying to solve the following two (separate) questions.

i) SinA= Cos2A

ii) Sin2A -1 = Cos2A

Where A must be presented in radians

Find A.

I'm afraid I'm a bit rusty on trigonometry as I have not done this in the 3 years I have been at Uni, so couldn't find many helpful resources on how to get started on this problem.


Think of how you can use double angle formulae here.
Reply 2
Avatar for B_9710
B_9710
18
For the first one use the double angle formula for cos and form a quadratic in sinA.
For the second one it will probably be helpful to cancel the -1 on the LHS so you may want to choose the appropriate double angle formula for cos2A.
(What course aware you doing at uni?)
Oh I see, I completely forgot about focusing on a quadratic in Sine! Perfect part i is all good.

For part 2 I am pretty sure I have done it right. So I use Cos2A = (CosA)^2 - 1 and remove the -1 as you have said. Then I rearrange to get tanA =0 and finally A = 0.

I did Maths and Economics, although throughout the course I hardly touched trigonometric identities.
Reply 4
Avatar for Zacken
Zacken
22
Original post by VincentCheung
Oh I see, I completely forgot about focusing on a quadratic in Sine! Perfect part i is all good.

For part 2 I am pretty sure I have done it right. So I use Cos2A = (CosA)^2 - 1 and remove the -1 as you have said. Then I rearrange to get tanA =0 and finally A = 0.

I did Maths and Economics, although throughout the course I hardly touched trigonometric identities.


Your penultimate step isn't quite right. You have sin⁡A=cos⁡2A\sin A = \cos^2 A, which doesn't get you tan⁡A=0\tan A = 0.

Instead, write everything in terms of sin⁡A\sin A again so you have sin⁡A−1=1−2sin⁡2A\sin A - 1 = 1 - 2\sin^2 A, now re-arrange to get a quadratic in sin⁡A\sin A.
Sorry it's so I used the double angle on both sides of the equation. So the and then after cancelling the -1 I was left with which I then divided by CosA and 2 on both sides to get .... But now I do see I've gone wrong as it should be 45 since dividing by CosA I put as 0 on the RHS, whereas it should be 1.
Reply 6
Avatar for B_9710
B_9710
18
Original post by VincentCheung
Sorry it's so I used the double angle on both sides of the equation. So the and then after cancelling the -1 I was left with which I then divided by CosA and 2 on both sides to get .... But now I do see I've gone wrong as it should be 45 since dividing by CosA I put as 0 on the RHS, whereas it should be 1.


You cannot divide by cosA since cosA=0 is a solution to the equation - in effect you are losing infinitely many solutions by dividing by cosA. You need to factorise.
Oh yes you're right.. So if I factorise then I get 0 is a solution along with the one I have found already? (A=45 degrees)
Original post by VincentCheung
Oh yes you're right.. So if I factorise then I get 0 is a solution along with the one I have found already? (A=45 degrees)

No, arccos(0) is 90 degrees.
Ah, another silly mistake.. But yes, thank you very much for your help!
Reply 10
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B_9710
18
Original post by VincentCheung
Ah, another silly mistake.. But yes, thank you very much for your help!


Also remember that there are infinitely many solutions unless the range of x values are restricted in the question.
If sin⁡x=1/2 \sin x= 1/2 for example, there will be infinitely many solutions of the form x=π/6+2nπ x=\pi /6 +2n\pi and also x=5π/6+2nπ,n∈Z x= 5\pi /6 +2n\pi, n \in \mathbb{Z} .

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