Is cos(4x)cos(x)-sin(4x)-sin(x)=0 ?

The answer to whether cos(4x)cos(x)-sin(4x)-sin(x)=0 is False

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prove cos(4x)cos(x)-sin(4x)-sin(x)=0

prove

cos (4 x) cos (x) - sin (4 x) - sin (x) = 0

False

Solution steps

cos (4 x) cos (x) - sin (4 x) - sin (x) = 0

Show that the two sides are not equal

For

cos (4 x) cos (x) - sin (4 x) - sin (x) = 0

plug in

x = 0

cos (4 \cdot 0) cos (0) - sin (4 \cdot 0) - sin (0) = 1

cos (4 \cdot 0) cos (0) - sin (4 \cdot 0) - sin (0)

Simplify:

4 \cdot 0 = 0

4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

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

cos (0) cos (0) - sin (0) - sin (0)

Add similar elements:

- sin (0) - sin (0) = - 2 sin (0)

= cos (0) cos (0) - 2 sin (0)

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

cos (0) cos (0)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (0) cos (0) = cos^{1 + 1} (0)

= cos^{1 + 1} (0)

Add the numbers:

1 + 1 = 2

= cos^{2} (0)

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

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

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

= 1^{2} - 2 \cdot 0

Simplify

= 1

The two sides are not equal

\Rightarrow False

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

sec (θ) + 1 = \frac{cos ( θ ) + 1}{cos ( θ )}

9 sin (7 π - x) = 9 sin (x)

\frac{cot ( x ) + tan ( x )}{cot ( x )} = sec^{2} (x)

1 + sin (4 a) + sin (2 a) = cos (2 a)

Is cos(4x)cos(x)-sin(4x)-sin(x)=0 ?
The answer to whether cos(4x)cos(x)-sin(4x)-sin(x)=0 is False