#1**+2 **

By the double-angle formula for sine:

\(\sin(2x)=2\sin(x)\cos(x)\\~\\ \sin(2x)=2\sin(x+\frac{\pi}{4}-\frac{\pi}{4})\cos(x+\frac{\pi}{4}-\frac{\pi}{4})\\~\\ \sin(2x)=2 \Big[\sin\Big((x+\frac{\pi}{4})-\frac{\pi}{4}\Big)\Big]\Big[\cos\Big((x+\frac{\pi}{4})-\frac{\pi}{4}\Big)\Big] \)

By the difference formulas for sine and cosine:

\( \sin(2x)=2 \Big[\sin( x+\frac{\pi}{4} )\cos(\frac{\pi}{4})-\cos(x+\frac{\pi}{4})\sin(\frac{\pi}{4})\Big]\Big[\cos(x+\frac{\pi}{4})\cos(\frac{\pi}{4})+\sin(x+\frac{\pi}{4})\sin(\frac{\pi}{4})\Big] \\~\\ \sin(2x)=2 \Big[\sin( x+\frac{\pi}{4} )(\frac{\sqrt2}{2})-\cos(x+\frac{\pi}{4})(\frac{\sqrt2}{2})\Big]\Big[\cos(x+\frac{\pi}{4})(\frac{\sqrt2}{2})+\sin(x+\frac{\pi}{4})(\frac{\sqrt2}{2})\Big] \\~\\ \sin(2x)=2 (\frac{\sqrt2}{2})\Big[\sin( x+\frac{\pi}{4} )-\cos(x+\frac{\pi}{4})\Big](\frac{\sqrt2}{2})\Big[\cos(x+\frac{\pi}{4})+\sin(x+\frac{\pi}{4})\Big] \\~\\ \sin(2x)=2 (\frac{\sqrt2}{2}) (\frac{\sqrt2}{2})\Big[\sin( x+\frac{\pi}{4} )-\cos(x+\frac{\pi}{4})\Big]\Big[\sin(x+\frac{\pi}{4})+\cos(x+\frac{\pi}{4})\Big] \\~\\ \sin(2x)=\Big[\sin( x+\frac{\pi}{4} )-\cos(x+\frac{\pi}{4})\Big]\Big[\sin(x+\frac{\pi}{4})+\cos(x+\frac{\pi}{4})\Big] \\~\\ \sin(2x)=\sin^2( x+\frac{\pi}{4} )-\cos^2(x+\frac{\pi}{4}) \)

By the Pythagorean identity:

\(\sin(2x)=(\,1-\cos^2(x+\frac{\pi}{4})\,)-\cos^2(x+\frac{\pi}{4})\\~\\ \sin(2x)=(\,1-m^2\,)-m^2\\~\\ \sin(2x)=1-2m^2\)

Guest Mar 14, 2019

#3**+3 **

given cos(π/4 + x) = m

what's the value of sin2x, in terms of m?

How about:

\(cos(π/4 + x) = m\\ cos(\pi/4)cosx - sin(\pi/4)sinx=m\\ \frac{cosx}{\sqrt2}-\frac{sinx}{\sqrt2}=m\\ cosx-sinx=\sqrt2\;m\\ \text{square both sides}\\ cos^2x+sin^2x-2cosxsinx=2m^2\\ 1-2cosxsinx=2m^2\\ -2cosxsinx=2m^2-1\\ 2cosxsinx=1-2m^2\\ sin(2x)=1-2m^2 \)

Melody Mar 15, 2019

#6**+1 **

thanks so much to both the users who posted answers!

In the second answer, I'm just confused about where the dividing by sqrt(2) comes from, since sin(pi/4) and cos(pi/4) equal sqrt(2) over 2, not just sqrt(2)

hearts123
Mar 15, 2019

#7**+2 **

Oh I thought there might have been a simpler way

Also, to hearts123, this might help answer your question:

\(\frac{\sqrt2}{2}=\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{(\sqrt2)^2}{2\sqrt2}=\frac{2}{2\sqrt2}=\frac{1}{\sqrt2}\)

Guest Mar 15, 2019

#9**+1 **

You had me confused for a moment, because I only ever think of them as 1/sqrt2

But guest is correct, they are equivalent.

I never remember the ratios like you obviously do, I just know the triangles that that they come fom.

\(\frac{\pi}{4}\; radians = 45^\circ\\~\\ cos45^\circ=sin45^\circ=\frac{1}{\sqrt2}\)

Melody
Mar 15, 2019