What is the value of sqrt(16sin^2(pi/4)+49cos^2(pi/4)) ?

The value of sqrt(16sin^2(pi/4)+49cos^2(pi/4)) is sqrt(65/2)

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sqrt(16sin^2(pi/4)+49cos^2(pi/4))

16 sin^{2} (\frac{π}{4}) + 49 cos^{2} (\frac{π}{4})

\frac{65}{2}

Decimal

5.70087 \dots

Solution steps

16 sin^{2} (\frac{π}{4}) + 49 cos^{2} (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

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}

= \frac{2}{2}

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{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}

= 16 (\frac{2}{2})^{2} + 49 (\frac{2}{2})^{2}

Simplify

16 (\frac{2}{2})^{2} + 49 (\frac{2}{2})^{2} : \frac{65}{2}

16 (\frac{2}{2})^{2} + 49 (\frac{2}{2})^{2}

Add similar elements:

16 (\frac{2}{2})^{2} + 49 (\frac{2}{2})^{2} = 65 (\frac{2}{2})^{2}

= 65 (\frac{2}{2})^{2}

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

= 65 \cdot \frac{1}{2}

Multiply

65 \cdot \frac{1}{2} : \frac{65}{2}

65 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 65}{2}

Multiply the numbers:

1 \cdot 65 = 65

= \frac{65}{2}

= \frac{65}{2}

= \frac{65}{2}

What is the value of sqrt(16sin^2(pi/4)+49cos^2(pi/4)) ?
The value of sqrt(16sin^2(pi/4)+49cos^2(pi/4)) is sqrt(65/2)