Is 3-6sin^2(x)=-3cos(2x) ?

The answer to whether 3-6sin^2(x)=-3cos(2x) is False

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 3-6sin^2(x)=-3cos(2x)

prove

3 - 6 sin^{2} (x) = - 3 cos (2 x)

False

Solution steps

3 - 6 sin^{2} (x) = - 3 cos (2 x)

Show that the two sides are not equal

For

3 - 6 sin^{2} (x) = - 3 cos (2 x)

plug in

x = 0

3 - 6 sin^{2} (0) = 3

3 - 6 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

= 3 - 6 \cdot 0^{2}

Simplify

= 3

- 3 cos (2 \cdot 0) = - 3

- 3 cos (2 \cdot 0)

Simplify

= - 3 cos (0)

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

= - 3 \cdot 1

Simplify

= - 3

The two sides are not equal

\Rightarrow False

(sin (\frac{5 \cdot \frac{π}{3}}{4})) = (sin (5 \frac{π}{12}))

tan (\frac{5 π}{12}) = tan (\frac{π}{4} + \frac{π}{6})

sin^{2} (θ) = csc^{2} (θ)

tan (x) (sec^{2} (x) - 1) = tan^{3} (x)

sin (x) = - 1