Double and half angle formulae

Hey, am stuck on a couple of questions.....

1) Given that tanθ = - 3/4 and that θ is obtuse, find the exact value of:
a) sec 2θ
b) cosec2θ
c) tan 2θ

I cant see how to draw the triangle if it has a right angle.... all I could think of was that its a 3,4,5 triangle and now im just totally stuck !

2) Obtain an expression for cos3θ in terms of cosθ and hence find the value of cos 3θ if cosθ=1/4

Well, I splid the cos 3θ up into 2θ andθ and used the double angle formula, but I just cant get it to be All cos. :frown:

3) Got a load of these proofs, Ive done most of them but theres a couple I canny do... ive tried doing one side and splitting it up and working from both ends and meeting in the middle - but nowt is working!

a) cos2A / (cosA-sinA) ≡ cosA + sinA how?!

b) tan2AsecA ≡ 2sinAsec2A

RIGHT, help with any of these is really really appreciated !

franks xxx

Reply 1

*bobo*

8

franks

Hey, am stuck on a couple of questions.....

1) Given that tanθ = - 3/4 and that θ is obtuse, find the exact value of:
a) sec 2θ
b) cosec2θ
c) tan 2θ

I cant see how to draw the triangle if it has a right angle.... all I could think of was that its a 3,4,5 triangle and now im just totally stuck !

2) Obtain an expression for cos3θ in terms of cosθ and hence find the value of cos 3θ if cosθ=1/4

Well, I splid the cos 3θ up into 2θ andθ and used the double angle formula, but I just cant get it to be All cos. :frown:

3) Got a load of these proofs, Ive done most of them but theres a couple I canny do... ive tried doing one side and splitting it up and working from both ends and meeting in the middle - but nowt is working!

a) cos2A / (cosA-sinA) ≡ cosA + sinA how?!

b) tan2AsecA ≡ 2sinAsec2A

RIGHT, help with any of these is really really appreciated !

franks xxx

3)
cos2A= cos² - sin² A
= (cos A- sin A)(cosA+sinA)
cos2A/(cosA-sinA)=cos A+sinA

Reply 2

BCHL85

12

Here are some fomulae
cos2A = cos²A - sin²A = 2cos²A - 1 = 1- 2sin²A
sin2A = 2sinAcosA
cos3A = 4cos³A - 3cosA
sin3A = 3sinA - 4sin³A
And if t = tan(A/2), sinA = 2t/(1+t²), cosA = (1-t²)/(1+t²)
I think you can use it to do the problems

Reply 3

Twiglet

OP

2

Yes, I know all those equations, apart from

cos3A = 4cos³A - 3cosA
sin3A = 3sinA - 4sin³A

which Im not allowed to use...
I have rearranged the formulae using these but I just cant get it to work right through to the answer :frown:

OP

can anyone help??

With the first question, you have done the right thing - drawing a 3,4,5 triangle. Ignore the negative sign for now.

For part A, you know that sec is equal to 1 / cos, yes? So, using your triangle so you can use the double angle formulae of cos 2A, and then find 1 / cos 2A. This will give you a fraction. Now, you was given in the question that the angle is obtuse - ie between 90 and 180 degrees. Think of your CAST diagram - only sin is positive between these values, so you know that sec of the angle must be negative.

Then you can use the same method for the other parts of the question.

The reason why the value of tan A was negative is purely due to the nature of the tan function; as A is obtuse, then the value of tan A must be negative.

Hope that makes sense!

Reply 6

Geminus

2

For the second question, you said you couldn't get "all cos". Without looking at it much closer (or getting to the stage where you have got to), I should think that you need to consider the identity sin^2 A + cos ^2 A = 1.

If you're still stuck with that one, I'll expand it out for you if you like :smile:

Reply 7

JaF

1

cos (2a+a) = cos(2a)cos(a) -sin(2a)sin(a)

cos(2a)=cos(a+a)=cos^2(a) - sin^2(a)
cos(2a)=cos^2(a)-(1-sin^2(a)) = 2cos^2(a) -1

sin(2a) = sin(a+a) = sin(a)cos(a) +cos(a)sin(a)
sin(2a)=2sin(a)cos(a)

subsitute those values into cos (2a+a) = cos(2a)cos(a) -sin(2a)sin(a) and use sin^2(a0 = 1-cos^2(a) to get rid of the sin^2.