What is the value of 2cos^2((3pi)/4)-1 ?

The value of 2cos^2((3pi)/4)-1 is 0

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2cos^2((3pi)/4)-1

2 cos^{2} (\frac{3 π}{4}) - 1

0

Solution steps

2 cos^{2} (\frac{3 π}{4}) - 1

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

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}

= - \frac{2}{2}

= 2 (- \frac{2}{2})^{2} - 1

Simplify

2 (- \frac{2}{2})^{2} - 1 : 0

2 (- \frac{2}{2})^{2} - 1

2 (- \frac{2}{2})^{2} = 1

2 (- \frac{2}{2})^{2}

(- \frac{2}{2})^{2} = \frac{1}{2}

(- \frac{2}{2})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{2}{2})^{2} = (\frac{2}{2})^{2}

= (\frac{2}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 2 ) ^{2}}{2 ^{2}}

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= \frac{2}{2 ^{2}}

Cancel the common factor:

2

= \frac{1}{2}

= 2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

= 0

sec (0.64)

2 - 2 sin (\frac{π}{4})

\frac{sin ( 3 5 ^{\circ} )}{1.33}

- 3 cos^{2} (π)

34 sin (3 0^{\circ})