Proof for Double Angle Formulae in a diagram

Masaru

Junior Member
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Sep 6, 2013
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59
Please see the attachment and have a look at the diagram for this question.

This is a question to prove double angle formulae using the diagram of a semi-circle of radius 1 unit.

As you can see, it says as follows:

In the diagram, the semi-circle has radius 1 unit, and angle PAB = theta.
Angle APO = theta {Triangle AOP is isosceles}
Angle PON = 2*theta {exterior angle of a triangle}

a) Find in terms of theta, the lengths of:

i) [OM]

sin (theta)

ii) [AM]
cos (theta)

iii) [ON]
cos (2*theta)

iv) [PN]
sin (2*theta)

So far so easy. But I have difficulty with the following question:

b) Use the triangle ANP and the lengths in a) to show that:

i) cos (theta) = sin (2*theta)/2*sin (theta)


From a) above,
sin (2*theta) = [PN], which is the opposite side from the angle PAB (theta) in the triangle ANP.
But I cannot identify 2*sin (theta), which is 2*[OM], in the triangle ANP.
Why double [OM]?
Why do we need to divide [PN] by double [OM] to get cos (theta)???

I would much appreciate it if someone can help me to make sense of this equation above from this diagram.

Thank you.
 

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Update: I have worked it out.

Sorry, but I have finally worked it out, so I do not need help any more.

My working is as follows:

The triangles ANP and AMO are actually similar.

So OM/AO = PN/AP

So sin (theta)/1 = sin (2*theta)/2*cos (theta)

So cos (2*theta) = sin (2*theta)/2*sin (theta) as required.
 
Please see the attachment and have a look at the diagram for this question.

This is a question to prove double angle formulae using the diagram of a semi-circle of radius 1 unit.

As you can see, it says as follows:

In the diagram, the semi-circle has radius 1 unit, and angle PAB = theta.
Angle APO = theta {Triangle AOP is isosceles}
Angle PON = 2*theta {exterior angle of a triangle}

a) Find in terms of theta, the lengths of:

i) [OM]

sin (theta)

ii) [AM]
cos (theta)

iii) [ON]
cos (2*theta)

iv) [PN]
sin (2*theta)

So far so easy. But I have difficulty with the following question:

b) Use the triangle ANP and the lengths in a) to show that:

i) cos (theta) = sin (2*theta)/2*sin (theta)


From a) above,
sin (2*theta) = [PN], which is the opposite side from the angle PAB (theta) in the triangle ANP.
But I cannot identify 2*sin (theta), which is 2*[OM], in the triangle ANP.
Why double [OM]?
Why do we need to divide [PN] by double [OM] to get cos (theta)???

I would much appreciate it if someone can help me to make sense of this equation above from this diagram.

Thank you.
In my opinion, it's a bit of a silly way to pose the question.

Rather than consider cos (theta), consider sin (theta) in triangle ANP.

That is sin (theta) = PN/AP = sin (2*theta) / 2 cos (theta) (Don't know why they just didn't ask for this)

Then make cos (theta) the subject.

Seems silly to me because you could answer the next question just as easily with sin (theta) the subject.

Let me know if that doesn't make sense.
 
Sorry, but I have finally worked it out, so I do not need help any more.

My working is as follows:

The triangles ANP and AMO are actually similar.

So OM/AO = PN/AP

So sin (theta)/1 = sin (2*theta)/2*cos (theta)

So cos (2*theta) = sin (2*theta)/2*sin (theta) as required.
No 2* here
 
In my opinion, it's a bit of a silly way to pose the question.

Rather than consider cos (theta), consider sin (theta) in triangle ANP.

That is sin (theta) = PN/AP = sin (2*theta) / 2 cos (theta) (Don't know why they just didn't ask for this)

Then make cos (theta) the subject.

Seems silly to me because you could answer the next question just as easily with sin (theta) the subject.

Let me know if that doesn't make sense.

By "next question" I meant (c).
 
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