solving trig identity: sinx = 1 - 2sin^2(pie/4 - x/2)

[darkness222]

i have a trig identity question that im stuck in it.

proof that:

sinx= 1-2sin^2(pie/4 - x/2)

so i started with the right side but then didnt get the same term on the left side.

waiting for your help

 

[Deleted member 4993]

Re: help with this trig identity?

i have a trig identity question that im stuck in it.

proof that:

sinx= 1-2sin^2(pie/4 - x/2)

so i started with the right side but then didnt get the same term on the left side.

waiting for your help

Please show us your work - even if you know it is wrong, so that we know where to begin to help you.

 

[galactus]

Re: help with this trig identity?

Anyone who spells \(\displaystyle {\pi}\) as 'pie' will be sentenced to hard labor picking fly poop out of pepper with boxing gloves. :roll: :wink:

 

[darkness222]

Re: help with this trig identity?

here's my work

R.S: 1-2(sin[pie/4-x/2)(sin[pie/4-x/2)

1-2 ((sin pi/4 * cos x/2) - (cos pi/4 * sin x/2 ))^2

1-2 ( root2/4 cosx - root2/4 sinx ) ^2

1-2 ( 1/8 cos^2x - 1/4 sinxcosx + 1/8 sin^2x)

1- 1/4cos^2x + 1/2sinxcosx - 1/4 sin^2 x

then i changed 1/4cos^2x + 1/4 sin^2 x to 1/4 since sin^2+cos^2=1

then its 3/4 + 3/2 sinxcosx

that was my solution for right side, but apparently it doesnt match the left side. so did i do something wrong in the steps?

 

[Deleted member 4993]

Re: help with this trig identity?

here's my work

R.S: 1-2(sin[pie/4-x/2)(sin[pie/4-x/2)

1-2 ((sin pi/4 * cos x/2) - (cos pi/4 * sin x/2 ))^2

1-2 ( root2/4 cosx - root2/4 sinx ) ^2<<<<< How's that!!!

1-2 ( 1/8 cos^2x - 1/4 sinxcosx + 1/8 sin^2x)

1- 1/4cos^2x + 1/2sinxcosx - 1/4 sin^2 x

then i changed 1/4cos^2x + 1/4 sin^2 x to 1/4 since sin^2+cos^2=1

then its 3/4 + 3/2 sinxcosx

that was my solution for right side, but apparently it doesnt match the left side. so did i do something wrong in the steps?

If I were to this problem I'll do it following way:

Let:

\(\displaystyle \theta \, = \, \frac{\pi}{4} \, - \, \frac{x}{2}\)

and

\(\displaystyle 2\theta \, = \, \frac{\pi}{2} \, - \, x\)

Then the right-hand-side becomes

\(\displaystyle 1 \, - \, 2\cdot \sin^2(\theta)\)

\(\displaystyle = \, \cos (2\theta)\)

Now continue......