What is the value of 20cos^2(15) ?

The value of 20cos^2(15) is 10((sqrt(3))/2+1)

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20cos^2(15)

20 cos^{2} (1 5^{\circ})

10 (\frac{3}{2} + 1)

Decimal

18.66025 \dots

Solution steps

20 cos^{2} (1 5^{\circ})

Rewrite using trig identities:

cos^{2} (1 5^{\circ}) = \frac{1 + cos ( 3 0 ^{\circ} )}{2}

cos^{2} (1 5^{\circ})

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

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

= \frac{1 + cos ( 2 \cdot 1 5 ^{\circ} )}{2}

Simplify

= \frac{1 + cos ( 3 0 ^{\circ} )}{2}

= 20 \cdot \frac{1 + cos ( 3 0 ^{\circ} )}{2}

Simplify

= 10 (cos (3 0^{\circ}) + 1)

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= 10 (\frac{3}{2} + 1)

arctan (\frac{7.9253}{22.356 - 11.184 - 7.061})

2 sin (1) cos (1)

cos (2 (\frac{5 π}{6}))

sin (\frac{π}{3} + \frac{π}{3})

arccos (\frac{2}{3 38})

What is the value of 20cos^2(15) ?
The value of 20cos^2(15) is 10((sqrt(3))/2+1)