Is 4sin^2(θ)+8cos(θ)=3 ?

The answer to whether 4sin^2(θ)+8cos(θ)=3 is False

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prove 4sin^2(θ)+8cos(θ)=3

prove

4 sin^{2} (θ) + 8 cos (θ) = 3

False

Solution steps

4 sin^{2} (θ) + 8 cos (θ) = 3

Show that the two sides are not equal

For

4 sin^{2} (θ) + 8 cos (θ) = 3

plug in

θ = 0

4 sin^{2} (0) + 8 cos (0) = 8

4 sin^{2} (0) + 8 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

= 4 \cdot 0^{2} + 8 \cdot 1

Simplify

= 8

The two sides are not equal

\Rightarrow False

sin^{2} (3 x) + cos^{2} (3 x) = 1

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