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

The answer to whether 1/2 cos(2x)=sin^2(x) is False

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

\frac{1}{2} cos (2 \cdot 0) = \frac{1}{2} (Decimal : 0.5)

\frac{1}{2} cos (2 \cdot 0)

Simplify:

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

\frac{1}{2} cos (0) = \frac{cos ( 0 )}{2}

\frac{1}{2} cos (0)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot cos ( 0 )}{2}

Multiply:

1 \cdot cos (0) = cos (0)

= \frac{cos ( 0 )}{2}

= \frac{cos ( 0 )}{2}

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= \frac{1}{2}

sin^{2} (0) = 0

sin^{2} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0^{2}

Simplify

= 0

The two sides are not equal

\Rightarrow False

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

\frac{4 + 2 sin ( θ )}{1 - sin ( θ )} = 3

\frac{sin ( 4 x )}{4} = 4 \frac{sin ( x ) cos ( x )}{4}

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

sin (- 3 0^{\circ}) = - sin (3 0^{\circ})