Is sqrt(1+sin(4a)+sin(2a))=-2cos(a) ?

The answer to whether sqrt(1+sin(4a)+sin(2a))=-2cos(a) is False

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prove sqrt(1+sin(4a)+sin(2a))=-2cos(a)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

a = 0

1 + sin (4 \cdot 0) + sin (2 \cdot 0) = 1

1 + sin (4 \cdot 0) + sin (2 \cdot 0)

Simplify

= 1 + 2 sin (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

= 1 + 2 \cdot 0

Simplify

= 1

- 2 cos (0) = - 2

- 2 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

= - 2 \cdot 1

Simplify

= - 2

The two sides are not equal

\Rightarrow False

(csc^{2} (θ)) (tan^{2} (θ)) - 1 = tan^{2} (θ)

cot (x) - cot (2 x) = csc (2 x)

\frac{csc ( θ ) + 1}{1 + sin ( θ )} = csc (θ)

\frac{tan ( a ) sin ( a )}{1 - cos ( a )} = sec (a) + 1

sec (A) csc (A) = csc (A) tan (A)

Is sqrt(1+sin(4a)+sin(2a))=-2cos(a) ?
The answer to whether sqrt(1+sin(4a)+sin(2a))=-2cos(a) is False