Proving Sin x + Sin 2x + Cos x + Cos 2x = 2√2cos(x/2)sin(3x/2+π/4)

[stfz]

Homework Statement

Prove that: $sin x + sin 2x + cos x + cos 2x = 2\sqrt{2} cos (\frac{x}{2}) sin(\frac{3x}{2}+\frac{\pi}{4})$

Homework Equations

We know all the double, compound, and half angle formulas.

The Attempt at a Solution

Taking on the RHS, we have

Expanding with half angle formula and compound angle formula

Hence we can cancel the sqrt(2) on the left, and replace sin/cos (pi/4) with exact values
And we have

If we expand the half angles in the right bracket

And then the sqrt(2) on the bottom can multiply to make 2, we can +/- them together

And cancel the twos.
Is it possible to cancel the +/- signs?
However, the graph of my result does not match that of the original.
For example, if I graph

As two graphs, one with a + and the other with a -, I find that the original graph

Is equal to the + graph in some intervals, and equal to the - graph in other intervals.

Please help! :)
Thanks

Stephen

 

[PeroK]

Homework Statement

Prove that: $sin x + sin 2x + cos x + cos 2x = 2\sqrt{2} cos (\frac{x}{2}) sin(\frac{3x}{2}+\frac{\pi}{4})$

Homework Equations

We know all the double, compound, and half angle formulas.

The Attempt at a Solution

Taking on the RHS, we have
View attachment 74954
Expanding with half angle formula and compound angle formula
View attachment 74954
Hence we can cancel the sqrt(2) on the left, and replace sin/cos (pi/4) with exact values
And we have
View attachment 74957
If we expand the half angles in the right bracket
View attachment 74958
And then the sqrt(2) on the bottom can multiply to make 2, we can +/- them together
View attachment 74959
And cancel the twos.
Is it possible to cancel the +/- signs?
However, the graph of my result does not match that of the original.
For example, if I graph
View attachment 74955
As two graphs, one with a + and the other with a -, I find that the original graph View attachment 74954
Is equal to the + graph in some intervals, and equal to the - graph in other intervals.

Please help! :)
Thanks

Stephen

Have you ever seen this formula? $Asin(x) + Bcos(x) = \sqrt{A^2+B^2}sin(x+α) \ \ (α = tan^{-1}(\frac{B}{A}))$

 

[stfz]

Hmm... Maybe that will work! Thanks for the suggestion, I will try it