Is 2sin^2(x)+4cos^2(x)=2-4sin^2(x) ?

The answer to whether 2sin^2(x)+4cos^2(x)=2-4sin^2(x) is False

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prove 2sin^2(x)+4cos^2(x)=2-4sin^2(x)

prove

2 sin^{2} (x) + 4 cos^{2} (x) = 2 - 4 sin^{2} (x)

False

Solution steps

2 sin^{2} (x) + 4 cos^{2} (x) = 2 - 4 sin^{2} (x)

Show that the two sides are not equal

For

2 sin^{2} (x) + 4 cos^{2} (x) = 2 - 4 sin^{2} (x)

plug in

x = 0

2 sin^{2} (0) + 4 cos^{2} (0) = 4

2 sin^{2} (0) + 4 cos^{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

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

= 2 \cdot 0^{2} + 4 \cdot 1^{2}

Simplify

= 4

2 - 4 sin^{2} (0) = 2

2 - 4 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

= 2 - 4 \cdot 0^{2}

Simplify

= 2

The two sides are not equal

\Rightarrow False

1 + sin (θ) = cos (θ)

1 + \frac{( tan ^{2} ( x ) )}{1 + sec ( x )} = sec (x)

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

csc^{2} (x) \cdot cos^{2} (x) = cot^{2} (x)

\frac{1}{sec ^{3} ( x ) cos ^{4} ( x )} = sec (x)

Is 2sin^2(x)+4cos^2(x)=2-4sin^2(x) ?
The answer to whether 2sin^2(x)+4cos^2(x)=2-4sin^2(x) is False