Question

HELP proving Trig Identity!

Answer 1

Prove \[\frac{1-2\sin ^2\left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}+2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)=\cos \left(x\right)\]

Answer 2

Hmmm....
\[1-2\sin^2x=\cos(2x)\]\[2\sin x\cos x = \sin (2x)\]Does that help a bit?

Answer 3

One more
\[\cos^2x - \sin^2x=\cos(2x)\]

Answer 4

FIrst, you need to simplify
\[1-2\sin^2x\] and\[2\sin(\frac{x}{2})\cos(\frac{x}{2})\]

Answer 5

does \[2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\]simplify to \[\frac{\sin(2x)\ }{ 2}\]?

Answer 6

No...

Answer 7

\[2\sin u\cos u= \sin(2u)\]
Take u = x/2
\[2sin(\frac{x}{2})cos(\frac{x}{2}) = ...?\]

Answer 8

so just sin(x)

Answer 9

Yes :)

Answer 10

After some simplification of the left side, what do you get so far?

Answer 11

\[\frac{\sin(x)\cos(x)+\sin^2x+\cos^2x}{\cos(x)+\sin(x)}\]

Answer 12

This looks a bit weird to me. Would you mind showing your workings?

Answer 13

\[\frac{\cos(2x)}{\cos(x)+sin(x)} + \sin(x)\]
then multiplied sin(x) by cos(x)+sin(x) to get common denominator.

Answer 14

Hmmm... Not \(cos^2x\)

My apology, maybe we need not simplify that much..
\[LS \]\[=\frac{1-2\sin ^2\left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}+2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\]\[=\frac{1-2\sin ^2\left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}+\sin(x)\]\[=\frac{1-2\sin ^2\left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}+\frac{\sin(x)(\sin(x)+\cos(x))}{\sin(x)+\cos(x)}\]\[=\frac{1-2\sin ^2\left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}+\frac{\sin^2(x)+\sin(x)\cos(x))}{\sin(x)+\cos(x)}\]\[=\frac{1-2\sin ^2\left(x\right)+\sin^2(x)+\sin(x)\cos(x)}{\sin(x)+\cos(x)}\]
Now, simplify the numerator, what do you get?

Answer 15

\[\frac{ 1-\sin^2(x)+\sin(x)\cos(x) }{ \sin(x) }\]

Answer 16

*denominator sin(x)+cos(x)

Answer 17

Yup. Recall:
\[1-\sin^2x = \cos^2x\]So, you'll get
\[\frac{ 1-\sin^2(x)+\sin(x)\cos(x) }{ \sin(x)+\cos(x) }\]\[=\frac{ \cos^2x+\sin(x)\cos(x) }{ \sin(x)+\cos(x) }\]Now, factorize the numerator, what do you get?

Answer 18

Sorry guys, but this can be done in fewer easier to understand steps

Answer 19

\[\frac{ 1-2\sin ^{2}x }{\cos x+\sin x }+2\sin \frac{ x }{2 }\cos \frac{ x }{2 }= \frac{ 1-\sin ^{2}x-\sin ^{2}x }{ \cos x+\sin x } +\sin x \]
\[\frac{ \cos ^{2}x-\sin ^{2}x }{\cos x+\sin x }+\sin x =\frac{ \left( \cos x+\sin x \right)\left( \cos x-\sin x \right) }{\cos x+\sin x } +\sin x \]
\[=\cos x-\sin x+\sin x=\cos x\]

Answer 20

well done surjit. thats how its done!