Is 1+0+1=-2cos(0)+4 ?

The answer to whether 1+0+1=-2cos(0)+4 is True

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prove 1+0+1=-2cos(0)+4

prove

1 + 0 + 1 = - 2 cos (0) + 4

True

Solution steps

1 + 0 + 1 = - 2 cos (0) + 4

Manipulating left side

1 + 0 + 1

Simplify

= 2

Manipulating right side

- 2 cos (0) + 4

Simplify

4 - 2 cos (0) : 2

4 - 2 cos (0)

2 cos (0) = 2

2 cos (0)

Simplify

cos (0) : 1

cos (0)

Use the following trivial identity:

cos (0) = 1

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

= 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

= 4 - 2

Simplify

= 2

= 2

We showed that the two sides could take the same form

\Rightarrow True

(1 - sin^{2} (y)) (1 + tan^{2} (y)) = 1

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

\frac{sin ( x ) + tan ( x )}{1 + sec ( x )} = tan (x)

sec (A) - cos (A) = tan (A) sin (A)

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