Question

Will give medal and fan!

Answer 1

How do I find out if this is an identity?

Answer 2

Okay what's sin2x

Answer 3

What's cos2x

Answer 4

Hint:

\(\sf\Large cos(2\alpha)=cos^2\alpha-sin^2\alpha\)

\(\sf\Large cos(2\alpha)=2cos^2\alpha-1 \)

\(\sf\Large cos(2\alpha)= 1-2sin^2\alpha \)

\(\sf\Large sin(2\alpha)=2~cos\alpha ~sin\alpha\)

Answer 5

Wouldn't the sin(a) and the cos(a) be cancelled out if you divide them?

Answer 6

Which \(\sf\Large cos(2\alpha)=?\) formula should we use?

Answer 7

The first one, right?

Answer 8

Nope.

Answer 9

But before we get to that, let's simplify the other fraction. And then chose which formula to use.

Answer 10

\(\sf\LARGE \frac{sin2\alpha}{sin\alpha}=?\)

Tell me what you get when you simplify it. Use the identity I gave you above :)

Answer 11

2 cosa sina=sin2a so would I do 2 cos(a)sin(a)/sin(a)? Or would taking sin(a) out equal 2?

Answer 12

yup, and divide sin(a) in the numerator and denominator to simplify it now.

Answer 13

So it's 2cos(a)?

Answer 14

exactly

Answer 15

and now for cos(2a)

We need to chose the eqaution that has only cos in it.

Answer 16

cos(2α)=2cos2α−1

Answer 17

correct

Answer 18

so now what does this equal?

\(\sf\Large\frac{2cos^2\alpha-1}{cos\alpha}=?\)

Hint:

\(\sf\Large\frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}\)

Answer 19

2cos^2a/ cosa -1/cosa?

Answer 20

@TheSmartOne

Answer 21

yup, and simplify that

Answer 22

Is it cosa^2?

Answer 23

Simplify this: