1-cos x

How do you rewrite this in terms of sin x?

sin^{2}x+cos^{2}x=1

Rearrange for cosx and then substitute that into

1-cosx

Reply 2

MSI_10

Original post by Robbie242

sin^{2}x+cos^{2}x=1

Rearrange for cosx and then substitute that into

1-cosx

Okay so

cos^2x=1-sin^2x

then

cosx=1-sinx

Thought square rooting here is not allowed? e.g

8^2+6^2=100

but

sqrt{100}=/=6+8

Reply 3

Robbie242

Original post by MSI_10

Okay so

cos^2x=1-sin^2x

then

cosx=1-sinx

Thought square rooting here is not allowed? e.g

8^2+6^2=100

but

sqrt{100}=/=6+8

Ahh but that's simply not true

cos^{2}x=1-sin^{2}x

is true
but

cosx=\pm\sqrt{1-sin^{2}x}

1-cosx=1-\pm\sqrt{1-sin^{2}x}

1-cosx=...

Reply 4

MSI_10

Original post by Robbie242

Ahh but that's simply not true

cos^{2}x=1-sin^{2}x

is true
but

cosx=\pm\sqrt{1-sin^{2}x}

1-cosx=1-\pm\sqrt{1-sin^{2}x}

1-cosx=...

I know it wasn't true that's why I pointed out it wasn't allowed.

What is it equal to then?

Reply 5

Robbie242

Original post by MSI_10

I know it wasn't true that's why I pointed out it wasn't allowed.

What is it equal to then?

oh are square roots not allowed in the answer? then I don't know how you'd get 1-cosx

1-cosx=1\mp \sqrt{1-sin^{2}x}

(the minus just changes the plus or minus around)

1-cosx=2sin^2(\frac{x}{2})

but that's not in terms of sinx

(edited 10 years ago)

Reply 6

Noble.

\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1-2\sin^2(x)

Use this identity with a slight alteration.

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