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

The answer to whether cos(x)+1(cos(x)-1)=cos^2(x)-1 is False

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

False

Solution steps

cos (x) + 1 \cdot (cos (x) - 1) = cos^{2} (x) - 1

Show that the two sides are not equal

For

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

plug in

x = 0

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

cos (0) + 1 \cdot (cos (0) - 1)

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

= 1 + 1 \cdot (1 - 1)

Simplify

= 1

cos^{2} (0) - 1 = 0

cos^{2} (0) - 1

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

= 1^{2} - 1

Simplify

= 0

The two sides are not equal

\Rightarrow False

sin (3 x) = 3 sin (x) - 4 sin (3 x)

cos (2 b + b) = 4 cos^{3} (b) - 3 cos (b)

2 cos^{2} (A) = 1 + cos (2 A)

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

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

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