How to do this, idk where to start.

[Soracyn11]

Any idea how to do this problem?

 

[Romsek]

What happens when you let [MATH]C=\pi - (A + B)[/MATH]and expand out the last term on the left using [MATH]\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)[/MATH]?

 

[Soracyn11]

Can you use that property if the pi is in the cos?

 

[Steven G]

In any triangle \(\displaystyle A + B + C =\pi\). So \(\displaystyle C = \pi- (A + B)\)

 

[Soracyn11]

Yes I got that, what do I do next?

 

[Soracyn11]

Replacing C or A or B to say c= pi +(A - B) doesn't help

 

[Steven G]

Replacing C or A or B to say c= pi +(A - B) doesn't help

If you did what you said then I am not surprised that it did not help. C = pi - (A+B) NOT C = pi + (A-B)

You just asked if it was valid to use C = pi - (A+B) and I said yes and explained why. Still not sure if it will help but then again Romsek is a clever person. Then again, the fact that they said that A, B and C are the angles in a triangles makes me think he is correct.

Now can you write Cos(A) using the half angle formula? How about for Cos(B)? SOMEHOW you need to convert the lhs into angles that involve A/2, B/2 and C/2 !!

 

[pka]

How can any of you read that image?
Soracyn11, why can can not simply type out the questions?

 

[Dr.Peterson]

I haven't contributed because I haven't solved the problem, but it would be appropriate to check out the problem.

First, here is a slightly more readable image, with the crotch cropped out and the contrast increased:



Now, I can read it either as

Proof: In a triangle ABC, [MATH]\csc A + \csc B + \csc C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}[/MATH]​

or as

Proof: In a triangle ABC, [MATH]\cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}[/MATH]​

Are those three cosecants, or cosines? The first looks like "cos", but the others look more like "csc" to me.

This is, of course, a good argument for typing a problem, perhaps supplemented by an image -- ideally an image of the problem in print.

 

[hoosie]

To confirm the question consider case of equilateral triangle where A,B and C are 60°.
LHS = 1/2 +1/2 +1/2
= 1½
RHS = 1 + 4(1/2 × 1/2 × 1/2)
= 1½
So it’s cos not cosec.

 

[hoosie]

Some guiding steps:
Show: 1 + 4sin(A/2)sin(B/2)sin(C/2) = cosA + cosB + cosC
A + B + C = 180°
⇒ A/2 + B/2 + C/2 = 90°
(A/2 + B/2 ) and C/2 are complementary angles as they add to give 90°.

Can you express sin(C/2) in the form of its complementary angle?
Expand your answer using cos(α + β) = cosαcosβ - sinαsinβ.
Substitute your expanded answer into expression below:
1 + 4sin(A/2)sin(B/2)sin(C/2)

Multiply out brackets and simplify using
2sin(α/2)cos(α/2) = sinα and 2sin²(α/2) = 1 - cosα
Also remember cosα = -cos(180° - α)

If you have any trouble following any of these steps post what you have done and then we can see where to help you.