Is sin(2θ)+2sin(θ)-cos(θ)=2 ?

The answer to whether sin(2θ)+2sin(θ)-cos(θ)=2 is False

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prove sin(2θ)+2sin(θ)-cos(θ)=2

prove

sin (2 θ) + 2 sin (θ) - cos (θ) = 2

False

Solution steps

sin (2 θ) + 2 sin (θ) - cos (θ) = 2

Show that the two sides are not equal

For

sin (2 θ) + 2 sin (θ) - cos (θ) = 2

plug in

θ = 0

sin (2 \cdot 0) + 2 sin (0) - cos (0) = - 1

sin (2 \cdot 0) + 2 sin (0) - cos (0)

Simplify

= 3 sin (0) - cos (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

= 3 \cdot 0 - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

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

\frac{csc ( x ) + 1}{cot ( x ) + cos ( x )} = sec (x)

\frac{sin ( - x ) cot ( x )}{sin ( \frac{π}{2} - x )} = 1

csc^{2} (\frac{θ}{2}) = \frac{2}{1 - cos ( θ )}

sec (x) cos (x) + tan^{2} (x) = sec^{2} (x)

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