Is prove-3cos^2(θ)+11=-5sin(θ)+6 ?

The answer to whether prove-3cos^2(θ)+11=-5sin(θ)+6 is False

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prove-3cos^2(θ)+11=-5sin(θ)+6

prove

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

False

Solution steps

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

Show that the two sides are not equal

For

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

plug in

θ = 0

- 3 cos^{2} (0) + 11 = 8

- 3 cos^{2} (0) + 11

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^{2} + 11

Simplify

= 8

- 5 sin (0) + 6 = 6

- 5 sin (0) + 6

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

= - 5 \cdot 0 + 6

Simplify

= 6

The two sides are not equal

\Rightarrow False

(1 - tan (x))^{2} = 1 - 2 tan (x) + tan^{2} (x)

csc (x) cos (x) tan (x) = 1

(\frac{cot ^{2} ( x )}{1 + cot ^{2} ( x )}) = cos^{2} (x)

\frac{1 - cot ( - θ )}{cot ( - θ )} = - tan (θ) - 1

cos (6 0^{\circ}) = 2 cos^{2} (3 0^{\circ}) - 1

Is prove-3cos^2(θ)+11=-5sin(θ)+6 ?
The answer to whether prove-3cos^2(θ)+11=-5sin(θ)+6 is False