What is the value of 5sin^2((3pi)/2)+3cos^2(4pi)-7tan(pi/4) ?

The value of 5sin^2((3pi)/2)+3cos^2(4pi)-7tan(pi/4) is 1

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5sin^2((3pi)/2)+3cos^2(4pi)-7tan(pi/4)

Solution

5 sin^{2} (\frac{3 π}{2}) + 3 cos^{2} (4 π) - 7 tan (\frac{π}{4})

Solution

1

Solution steps

5 sin^{2} (\frac{3 π}{2}) + 3 cos^{2} (4 π) - 7 tan (\frac{π}{4})

cos (4 π) = cos (0)

cos (4 π)

Rewrite

4 π

2 π \cdot 2 + 0

= cos (2 π 2 + 0)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 2 + 0) = cos (0)

= cos (0)

= 5 sin^{2} (\frac{3 π}{2}) + 3 cos^{2} (0) - 7 tan (\frac{π}{4})

Rewrite using trig identities:

sin^{2} (\frac{3 π}{2}) = \frac{1 - cos ( π )}{2}

sin^{2} (\frac{3 π}{2})

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

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

= \frac{1 - cos ( 2 \cdot \frac{3 π}{2} )}{2}

Simplify:

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

Multiply fractions:

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

= \frac{3 π 2}{2}

Cancel the common factor:

2

= 3 π

= \frac{1 - cos ( 3 π )}{2}

cos (3 π) = cos (π)

cos (3 π)

Rewrite

3 π

2 π + π

= cos (2 π + π)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + π) = cos (π)

= cos (π)

= \frac{1 - cos ( π )}{2}

= 5 \cdot \frac{1 - cos ( π )}{2} + 3 cos^{2} (0) - 7 tan (\frac{π}{4})

5 \cdot \frac{1 - cos ( π )}{2} + 3 cos^{2} (0) - 7 tan (\frac{π}{4}) = \frac{5 ( 1 - cos ( π ) ) + 6 cos ^{2} ( 0 ) - 14 tan ( \frac{π}{4} )}{2}

5 \cdot \frac{1 - cos ( π )}{2} + 3 cos^{2} (0) - 7 tan (\frac{π}{4})

Multiply

5 \cdot \frac{1 - cos ( π )}{2} : \frac{5 ( - cos ( π ) + 1 )}{2}

5 \cdot \frac{1 - cos ( π )}{2}

Multiply fractions:

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

= \frac{( 1 - cos ( π ) ) \cdot 5}{2}

= \frac{5 ( - cos ( π ) + 1 )}{2} + 3 cos^{2} (0) - 7 tan (\frac{π}{4})

Convert element to fraction:

3 cos^{2} (0) = \frac{3 cos ^{2} ( 0 ) 2}{2}, 7 tan (\frac{π}{4}) = \frac{7 tan ( \frac{π}{4} ) 2}{2}

= \frac{( 1 - cos ( π ) ) \cdot 5}{2} + \frac{3 cos ^{2} ( 0 ) \cdot 2}{2} - \frac{7 tan ( \frac{π}{4} ) \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{( 1 - cos ( π ) ) \cdot 5 + 3 cos ^{2} ( 0 ) \cdot 2 - 7 tan ( \frac{π}{4} ) \cdot 2}{2}

(1 - cos (π)) \cdot 5 + 3 cos^{2} (0) \cdot 2 - 7 tan (\frac{π}{4}) \cdot 2 = 5 (1 - cos (π)) + 6 cos^{2} (0) - 14 tan (\frac{π}{4})

(1 - cos (π)) \cdot 5 + 3 cos^{2} (0) \cdot 2 - 7 tan (\frac{π}{4}) \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 5 (- cos (π) + 1) + 6 cos^{2} (0) - 7 \cdot 2 tan (\frac{π}{4})

Multiply the numbers:

7 \cdot 2 = 14

= 5 (- cos (π) + 1) + 6 cos^{2} (0) - 14 tan (\frac{π}{4})

= \frac{5 ( - cos ( π ) + 1 ) + 6 cos ^{2} ( 0 ) - 14 tan ( \frac{π}{4} )}{2}

= \frac{5 ( 1 - cos ( π ) ) + 6 cos ^{2} ( 0 ) - 14 tan ( \frac{π}{4} )}{2}

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

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

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= \frac{5 ( 1 - ( - 1 ) ) + 6 \cdot 1 ^{2} - 14 \cdot 1}{2}

Simplify

= 1

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Frequently Asked Questions (FAQ)

What is the value of 5sin^2((3pi)/2)+3cos^2(4pi)-7tan(pi/4) ?
The value of 5sin^2((3pi)/2)+3cos^2(4pi)-7tan(pi/4) is 1