# Thread: Trigonometric Identities and equations

1. ## Trigonometric Identities and equations

Simplify each expression.

5 c) 1-2sin^2 3π/8

I got 1-2sin^2 67.5

Cos2Θ= 1-2sin^2 67.5

I dont know exactly what to do from there...

2. Originally Posted by Skoz
Simplify each expression.

5 c) 1-2sin^2 3π/8

I got 1-2sin^2 67.5

Cos2Θ= 1-2sin^2 67.5

I dont know exactly what to do from there...
Leave it in terms of $\pi$ otherwise you will never solve it

$1-2sin^2 \left(\frac{3\pi}{8}\right) = cos \left(\frac{3\pi}{4}\right)$ (because $2 \times \frac{3}{8} = \frac{3}{4}$)

$cos \left(\frac{3\pi}{4}\right) = cos \left(\pi - \frac{\pi}{4}\right)$

To solve use the angle difference formula: $cos(A-B) = cosAcosB+sinAsinB$

Spoiler:
Exact values of sin and cos at the two above values:

$cos(\pi) = -1$
$sin(\pi) = 0$

$cos \left(\frac{\pi}{4}\right) = sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2}$

3. where does the 3pi/4 come from?

and how does cos (3pi/4)= (pi-pi/4) where did the 3 go?

4. Originally Posted by Skoz

where does the 3pi/4 come from?

and how does cos (3pi/4)= (pi-pi/4) where did the 3 go?
I used the double angle formula for cos:

$1-2sin^2 \theta = cos(2 \theta)$

Therefore $1-2sin^2 \left(\frac{3\pi}{8}\right) = cos \left(2 \cdot \frac{3\pi}{8}\right)$

$\frac{3\pi}{4}$ comes from simplifying the cos term

$2 \times \frac{3\pi}{8} = \frac{6\pi}{8} = \frac{3\pi}{4}$

The 3 is turned into a 4-1. If you imagine $\pi = \frac{4\pi}{4}$ then we see that
$\frac{4\pi}{4}-\frac{\pi}{4} = \frac{3\pi}{4}$

I used $\pi$and $\frac{\pi}{4}$ because their exact values are known