C3 Chapter7 Trig (Edexcel)

So I'm working through the first exercise in ch. 7 Ex 7A of C3, and I've been able to do the questions fine up until question 13. I thought I found a breakthrough, but much to my dissapointment the values I end up with are not the same as those in the book, and I don't think they are errors since for the other parts of 13 I am also unable to find the given values.

The question is as follows (sorry I'm not sure how to use latex yet):

3Costheta = 2Sin(theta + 60)

I first wrote the second expression using the trig identity SinACosB + CosASinB, then simplified the cos' and sin's and took the cos' to one side resulting with:

3Costheta - rt3.Costheta = Sintheta

I then squared both sides, wrote sin^2 as cos^2, collected like terms and ended with a quadratic in the form:

13Cos^2theta - 6.rt3.Costheta - 1 = 0

Solving for theta I then get 27.6, 332.4, 95 & 275. However the book gives values of 51.7 & 231.7.

I am unable to see the flaw in my working or method, though I do suspect that my method is not the intended one especially as using the same method I can't get the following questions correct. Can someone please help lead me across the right track?

Reply 1

username1230506

1

it shouldnt be 13cos^2(theta)

Reply 2

usycool1

19

Original post by TheFuture001

So I'm working through the first exercise in ch. 7 Ex 7A of C3, and I've been able to do the questions fine up until question 13. I thought I found a breakthrough, but much to my dissapointment the values I end up with are not the same as those in the book, and I don't think they are errors since for the other parts of 13 I am also unable to find the given values.

The question is as follows (sorry I'm not sure how to use latex yet):

3Costheta = 2Sin(theta + 60)

I first wrote the second expression using the trig identity SinACosB + CosASinB, then simplified the cos' and sin's and took the cos' to one side resulting with:

3Costheta - rt3.Costheta = Sintheta

I then squared both sides, wrote sin^2 as cos^2, collected like terms and ended with a quadratic in the form:

13Cos^2theta - 6.rt3.Costheta - 1 = 0

When solving an equation, be careful when you square both sides (btw, I think you squared the LHS incorrectly too :tongue:

) - you'll end up with "extraneous" solutions.

Instead, divide everything by cos theta after you get the bit in bold. :smile:

Reply 3

TheFuture001

OP

12

Original post by usycool1

When solving an equation, be careful when you square both sides (btw, I think you squared the LHS incorrectly too :tongue:

) - you'll end up with "extraneous" solutions.

Instead, divide everything by cos theta after you get the bit in bold. :smile:

Hmm, interesting, I'll give it a go. And could you explain what you mean by "extraneous"?

Reply 4

usycool1

19

Original post by TheFuture001

Hmm, interesting, I'll give it a go. And could you explain what you mean by "extraneous"?

Let me know if you need more help. :yep:

They're extra, invalid solutions. Have a look at this: http://mathmistakes.info/facts/AlgebraFacts/learn/ctm/extra.html :smile:

Reply 5

TheFuture001

OP

12

Haha thanks a lot mate, this question was really driving me through the wall, I should be able to do the rest now. Cheers!! :biggrin:

Original post by usycool1

They're extra, invalid solutions. Have a look at this: http://mathmistakes.info/facts/AlgebraFacts/learn/ctm/extra.html

Ohh, I did not know about this, most probably the reason the following questions contained right and wrong answers. Thanks, I'll take note of this. :biggrin:

(edited 10 years ago)

C3 Chapter7 Trig (Edexcel)

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