Trig identity: verify cos^2(x) - sin^2(x) = 2cos^2(x) - 1

Just note that \(\displaystyle \sin ^2 (x) + \cos ^2 (x) = 1\) and make a substitution.
 
TRIG Equation

Tring to verify Identity cos^2x-sin^2x=2cos^2x-1. I have tried substituting
sin^2+cos^2=1 , but I am not havig any luck.
 
Timothy said:
I have tried substituting sin^2+cos^2=1 , but I am not havig any luck.
Please reply showing how you did the substitution and what steps you took after that.

Thank you.

Eliz.
 
Trig Equation

One of my problems is , I have ( cos^2x-sin^2x ), on one side of the equal sign.
And I am not sure how to substitute (sin^2 + cos^2=1) sence the equation has a
negative sin.
My other problem is, if I try to simplify the other side of the equation , (2cos^2x-1)
I am not sure what to do with the 2 in front of the cos.
 
Just use the identity \(\displaystyle sin^{2}(x)=1-cos^{2}(x)\)

\(\displaystyle cos^{2}(x)-(1-cos^{2}(x))\)

See it now?.
 
Trig Equation

I see how you got , sin^2=1-cos^2 from sin^2+cos^2=1. But if you are
telling me to replace sin^2=1-cos^2 with the original equation cos^2-sin^2,
I don't under stand how we can do this.

Also, I am not sure were we are getting the cos^2-(1-cos^2) from the previous
reply.
I am sorry for the trouble. It takes me awhile to get this stuff.
I will work on it again tomarrow.

Thanks so much for your help . Tim
 
Re: Trig identity: verify cos^2(x) - sin^2(x) = 2cos^2(x) -

Timothy said:
Verify the Identity:

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

Ok...I think you "get" this (from the fundamental identities):

sin<SUP>2</SUP>x + cos<SUP>2</SUP>x = 1

Subtract cos<SUP>2</SUP>x from both sides of the above statement to get

sin<SUP>2</SUP> x = 1 - cos<SUP>2</SUP>

Now, substitute (1 - cos<SUP>2</SUP>) for sin<SUP>2</SUP>in the left side of this statement (the GIVEN in your problem):

cos<SUP>2</SUP> - (1 - cos<SUP>2</SUP>) = 2 cos<SUP>2</SUP> - 1

Simplify the left side:

cos<SUP>2</SUP> - 1 + cos<SUP>2</SUP> = 2 cos<SUP>2</SUP> - 1

2 cos<SUP>2</SUP> - 1 = 2 cos<SUP>2</SUP> - 1

Ok?
 
Trig Equation

I see it now. Sorry it took me so long. I have been out of school for 30 years,
and I am taking some collage classes for work.

Thank you very much.
Tim
 
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