Is 3cos(θ)=sqrt(3)sin(-θ) ?

The answer to whether 3cos(θ)=sqrt(3)sin(-θ) is False

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prove 3cos(θ)=sqrt(3)sin(-θ)

prove

3 cos (θ) = 3 sin (- θ)

False

Solution steps

3 cos (θ) = 3 sin (- θ)

Show that the two sides are not equal

For

3 cos (θ) = 3 sin (- θ)

plug in

θ = 0

3 cos (0) = 3

3 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

= 3 \cdot 1

Simplify

= 3

3 sin (- 0) = 0

3 sin (- 0)

Rewrite using trig identities:

sin (- 0) = 0

sin (- 0)

Simplify

= 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

= 0

= 3 \cdot 0

Simplify

= 0

The two sides are not equal

\Rightarrow False

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

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

\frac{sin ( 3 a )}{sin ( a )} = 2 cos (2 a) + 1

\frac{1}{sin ( x ) cos ( x )} = \frac{sec ( x )}{sin ( x )}

sin (2 x) + cos (2 x) = 1 + 2 sin (x) cos (x)

Is 3cos(θ)=sqrt(3)sin(-θ) ?
The answer to whether 3cos(θ)=sqrt(3)sin(-θ) is False