# Help with trig proofs.

• Apr 4th 2007, 02:12 PM
Help with trig proofs. [edit: one remaining]
I'm stumped on three problems that were assigned on a math project for Algebra II/Trig.

[SOLVED] 1) Express the following function as a single circular sinusoid.
f(x)=sqrt(3)*sin(2x)+cos(2x)

2) Prove that if y1=a1*sin(bx) and y2=a2*cos(bx), then A is an element of the real number and B is an element of the real number such that a1*sin(bx)+a2*cos(bx)=Asin(k(x+B)). A should be in terms of a1 and a2, and there should be only one expression for B. Also, state what the constant k is equal to.

[SOLVED, YAY!] 3) Prove that:
(1/2)+cos(x)+cos(2x)+cos(3x)......+cos(nx)=

sin((n+1/2)x)
---------------
2sin(x/2)

I apologize for the bad formatting (don't know how to get MathType into the forum. )

Any help would be greatly appreciated.
• Apr 4th 2007, 05:22 PM
Soroban

. . The solution is a strange one . . .

Quote:

1) Express the following function as a single circular sinusoid:
. . f(x) .= .√3·sin(2x) + cos(2x)

Divide both sides by 2:

. . ˝·f(x) .= .˝·√3·sin(2x) + ˝·cos(2x)

We know that: .cos(30°) = ˝√3 .and .sin(30°) = ˝

So we have: .˝·f(x) .= .cos(30°)·sin(2x) + sin(30°)·cos(2x)

. . . . . - . . . - . . . . .= .sin(2x)·cos(30°) + cos(2x)·sin(30°)

From the compound-angle identity: .sin(A + B) .= .sin(A)·cos(B) + cos(A)·sin(B)

. . we have: .˝·f(x) .= .sin(2x + 30°)

Therefore: .f(x) .= .2·sin(2x + 30°)

• Apr 4th 2007, 05:49 PM
ThePerfectHacker
Quote:

3) Prove that:
(1/2)+cos(x)+cos(2x)+cos(3x)......+cos(nx)=

sin((n+1/2)x)
---------------
2sin(x/2)

I apologize for the bad formatting (don't know how to get MathType into the forum. )

Any help would be greatly appreciated.

This is not true.

Try n=1.
• Apr 4th 2007, 08:03 PM
Thanks for the help!

@Soroban:
Quote:

The solution is a strange one . . .
my teacher adores tricky/strange problems... It's extremely frustrating for us, the students.
Thanks for the answer, been staring at it and trying to work out something for a long time. Would've never thought of dividing both sides by 2.

@ThePerfectHacker: I did try n=1, and graphing it, it's the same:
y1= (1/2) + cos(x)
y2=sin((1+1/2)x)/(2sin(x/2))
both give same graph. Of course, y2 doesn't work when 2sin(x/2)=0. But otherwise, it worked =/.

EDIT: SOLVED, just #2 remaining