Is sin^2(2θ)=(1-cos(4θ))/2 ?

The answer to whether sin^2(2θ)=(1-cos(4θ))/2 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin^2(2θ)=(1-cos(4θ))/2

prove

sin^{2} (2 θ) = \frac{1 - cos ( 4 θ )}{2}

True

Solution steps

sin^{2} (2 θ) = \frac{1 - cos ( 4 θ )}{2}

Manipulating right side

\frac{1 - cos ( 4 θ )}{2}

Rewrite using trig identities

\frac{1 - cos ( 4 θ )}{2}

= \frac{1 - cos ( 2 \cdot 2 θ )}{2}

Use the Double Angle identity:

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

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

Simplify

\frac{1 - ( 1 - 2 sin ^{2} ( 2 θ ) )}{2} : sin^{2} (2 θ)

\frac{1 - ( 1 - 2 sin ^{2} ( 2 θ ) )}{2}

Expand

1 - (1 - 2 sin^{2} (2 θ)) : 2 sin^{2} (2 θ)

1 - (1 - 2 sin^{2} (2 θ))

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

- (1 - 2 sin^{2} (2 θ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (2 θ))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (2 θ)

= 1 - 1 + 2 sin^{2} (2 θ)

1 - 1 = 0

= 2 sin^{2} (2 θ)

= \frac{2 sin ^{2} ( 2 θ )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin^{2} (2 θ)

= sin^{2} (2 θ)

= sin^{2} (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{tan ( x )} = arctan (x)

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

sec (- x) = \frac{1}{cos ( x )}

csc (x) = 2

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

Is sin^2(2θ)=(1-cos(4θ))/2 ?
The answer to whether sin^2(2θ)=(1-cos(4θ))/2 is True