Question

Am I on the right track????
2 Sin^2 (45deg - theta)
becomes...
Sin^2 theta (sin (45deg - theta)
which becomes...
Sin^2 theta (sin 45 cos theta - cos 45 sin theta)?

Answer 1

\(\large \rm \sin^2( x)\) is same as \(\large \rm \left(\sin( x)\right)^2\)

Answer 2

So then the (45deg - theta) would be squared?

Answer 3

\[\large \rm 2\sin^2(45 - \theta) = 2\left[\sin (45-\theta)\right]^2 \]

Answer 4

\[\large \rm = 2\left[\sin (45) \cos(\theta) - \cos(45) \sin(\theta)\right]^2 \]

Answer 5

\[\large \rm = 2\left[\frac{1}{\sqrt{2}} \cos(\theta) - \frac{1}{\sqrt{2}} \sin(\theta)\right]^2 \]

Answer 6

\[\large \rm = 2\frac{1}{2}\left[\cos(\theta) - \sin(\theta)\right]^2 \]

Answer 7

\[\large \rm = \left[\cos(\theta) - \sin(\theta)\right]^2 \]

Answer 8

use (a-b)^2 = a^2+b^2-2ab and simplify

Answer 9

one moment let me see if i can simplify...

Answer 10

so it just becomes Cos^2 theta + Sin^2 theta?

Answer 11

\[\large \rm \left[\cos(\theta) - \sin(\theta)\right]^2 = \cos^2(\theta) + \sin^2(\theta) - 2\sin(\theta)\cos(\theta)\]

Answer 12

recall the identity : sin^2 + cos^2 = 1

Answer 13

\[\large \rm \begin{align} \left[\cos(\theta) - \sin(\theta)\right]^2 &= \cos^2(\theta) + \sin^2(\theta) - 2\sin(\theta)\cos(\theta)\\~\\&=1 - \sin (2\theta)\end{align}\]

Answer 14

oh yes. I see, but one question im confused on, is how did you get the 1/2 back in step 4?

Answer 15

\[\large \rm 2\left[\frac{1}{\sqrt{2}} \cos(\theta) - \frac{1}{\sqrt{2}} \sin(\theta)\right]^2\]

after this step ?

Answer 16

yes

Answer 17

factor 1/sqrt(2) from both terms

Answer 18

\[\large \rm 2\left[\frac{1}{\sqrt{2}} (\cos(\theta) - \sin(\theta))\right]^2\]

Answer 19

but isn't the sin and cos of 1/radical 2 = radical 2/2?

Answer 20

oh my gosh , i m an idiot. of course. thank you

Answer 21

I appreciate your time.