Is cos(θ)=-(sqrt(3))/4 ?

The answer to whether cos(θ)=-(sqrt(3))/4 is False

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prove cos(θ)=-(sqrt(3))/4

prove

cos (θ) = - \frac{3}{4}

False

Solution steps

cos (θ) = - \frac{3}{4}

Show that the two sides are not equal

For

cos (θ) = - \frac{3}{4}

plug in

θ = 0

cos (0) = 1

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

= 1

The two sides are not equal

\Rightarrow False

sin (75) = sin (\frac{150}{2})

(sec^{2} (y)) = \frac{1}{cos ^{2} ( y )}

cos (4 x) cos (x) - sin (4 x) - sin (x) = 0

2 cos (2 x) = 2 sin (x^{2}) - 1

sec (θ) + 1 = \frac{cos ( θ ) + 1}{cos ( θ )}