Trigonometric Identities and equations

**Skoz** · December 6th 2009, 01:45 PM

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...

**e^(i*pi)** · December 6th 2009, 02:01 PM

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:

**Skoz** · December 6th 2009, 02:11 PM

where does the 3pi/4 come from?

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

**e^(i*pi)** · December 7th 2009, 09:49 AM

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

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