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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

= 1 + cos (2 x) - (cos (2 x) - 1)

1 + cos (2 x) - (cos (2 x) - 1) = 2

1 + cos (2 x) - (cos (2 x) - 1)

- (cos (2 x) - 1) : - cos (2 x) + 1

- (cos (2 x) - 1)

Distribute parentheses

= - (cos (2 x)) - (- 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos (2 x) + 1

= 1 + cos (2 x) - cos (2 x) + 1

Simplify

1 + cos (2 x) - cos (2 x) + 1 : 2

1 + cos (2 x) - cos (2 x) + 1

Add similar elements:

cos (2 x) - cos (2 x) = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

= 2

= 2

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{1}{1 + sin ( x )} + \frac{1}{1 + csc ( x )} = 1

p ro v e sin^{2} (θ) cos^{2} (θ) = 1

p ro v e 1 - 2 cos^{2} (θ) + cos^{4} (θ) = sin^{4} (θ)

p ro v e csc^{2} (θ) sec^{2} (θ) - csc^{2} (θ) = sec^{2} (θ)

p ro v e tan (a) + cot (a) = \frac{csc ( a )}{cos ( a )}

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