Is 1+sin^2(x)=2sin^2(x)+cos^2(x) ?

The answer to whether 1+sin^2(x)=2sin^2(x)+cos^2(x) is True

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prove 1+sin^2(x)=2sin^2(x)+cos^2(x)

prove

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

True

Solution steps

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

Manipulating right side

2 sin^{2} (x) + cos^{2} (x)

Rewrite using trig identities

2 sin^{2} (x) + cos^{2} (x)

Use the Pythagorean identity:

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

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

= 2 sin^{2} (x) + 1 - sin^{2} (x)

Simplify

2 sin^{2} (x) + 1 - sin^{2} (x) : sin^{2} (x) + 1

2 sin^{2} (x) + 1 - sin^{2} (x)

Group like terms

= 2 sin^{2} (x) - sin^{2} (x) + 1

Add similar elements:

2 sin^{2} (x) - sin^{2} (x) = sin^{2} (x)

= sin^{2} (x) + 1

= sin^{2} (x) + 1

= sin^{2} (x) + 1

= 1 + sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

csc (x) = \frac{1}{sec ( x )}

\frac{1 + sin ( α )}{cos ( α )} = \frac{cos ( α )}{1 - sin ( α )}

2 csc^{2} (y) = \frac{1}{1 - cos ( y )} + \frac{1}{1 + cos ( y )}

- \frac{sin ( x )}{1 - cos ( x )} = - csc (x) - cot (x)

(tan (b) + sec (b)) (1 - sin (b)) = cos (b)

Is 1+sin^2(x)=2sin^2(x)+cos^2(x) ?
The answer to whether 1+sin^2(x)=2sin^2(x)+cos^2(x) is True