In this discussion, we are going to prove the double angle trig identities for sine, cosine, and tangent; and we will do this using the sum formulas for sine, cosine, and tangent.
Here is an outline of this this discussion:
▪️Proving sin(2x) = 2sinxcosx
▪️Proving cos(2x) = cos^2x-sin^2x
▪️Proving cos(2x) = 2cos^2x-1
▪️Proving cos(2x) = 1-2sin^2x
▪️Proving tan(2x) = 2tanx/(1-tan^2x)
Let’s begin!
Proving sin(2x) = 2sinxcosx
We know by the sum formula that
sin(a+b)
= sinacosb +cosasinb.
So,
sin(x+x)
= sinxcosx + cosxsinx
= 2sinxcosx.
Thus,
sin(x+x) = 2sinxcosx, which means that
sin(2x) = 2sinxcosx
If you are interested in proving the sum identities, you can go to the following discussions:
Deriving the Sum and Difference formulas for Sine and Cosine.
Proving the Sum and Difference Formulas for Tangent and Cotangent
Otherwise, let’s continue.
Proving cos(2x) = cos^2x-sin^2x
We know by the sum formula for cosine that
cos(a+b)
= cosacosb-sinasinb.
So,
cos(x+x)
= cosxcosx – sinxsinx
= cos^2x – sin^2x.
Thus,
cos(x+x) = cos^2x-sin^2x,
which means that
cos(2x) = cos^2x-sin^2x.
Proving cos(2x) = 2cos^2x-1
We have shown that
cos(2x) = cos^2x-sin^2x.
Replacing sin^2x with 1-cos^2x, we have
cos(2x)
= cos^2x-sin^2x
= cos^2x – (1-cos^2x)
= cos^2x – 1 +cos^2x
= 2cos^2x -1.
Thus,
cos(2x) = 2cos^2x-1
Proving cos(2x) = 1-2sin^2x
We have shown that
cos(2x) = cos^2x-sin^2x.
Replacing cos^2x with 1-sin^2x, we have
cos(2x)
= cos^2x-sin^2x
= (1-sin^2x) – sin^2x
= 1 – sin^2x – sin^2x
= 1 – 2sin^2x. Thus,
cos(2x) = 1-2sin^2x.
Proving tan(2x) = 2tanx/(1-tan^2x)
We know by the sum formula for tangent that
tan(a+b)
= (tana + tanb)/(1-tanatanb).
So,
tan(x+x)
= (tanx + tanx)/(1-tanxtanx)
= 2tanx/(1-tan^2x).
Thus,
tan(x+x) = 2tanx/(1-tan^2x),
which means that
tan(2x) = 2tanx/(1-tan^2x).
We have proved the Double-Angle Formula for Sine, Cosine, and Tangent!
Thank you for reading!
I hope that I can assist you on your math journey!
Sincerely,
➕MathAngel369➕
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