If $a = 6, b = 8$ and $c = 10$ then $cos 4 C + sin 4 C =$

\frac{4}{5}

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\frac{2}{3}

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1

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\frac{3}{5}

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Solution

The correct option is C $1$
$c^{2} = a^{2} + b^{2}$

$10^{2} = 6^{2} + 8^{2}$

So triangle is right angle at C

So $∠ c = \frac{π}{2}$

$= cos 4 (\frac{π}{2}) + sin (4 \times \frac{π}{2})$

$= cos 2 π + sin 2 π$

$= 1 + 0$

$= 1$

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If a=6, b=8 and c=10 then cos4C+sin4C=

The correct option is C 1c2=a2+b2102=62+82So triangle is right angle at CSo ∠c=π2=cos4(π2)+sin(4×π2)=cos2π+sin2π=1+0=1

If $a = 6, b = 8$ and $c = 10$ then $cos 4 C + sin 4 C =$

The correct option is C $1$
$c^{2} = a^{2} + b^{2}$

$10^{2} = 6^{2} + 8^{2}$

So triangle is right angle at C

So $∠ c = \frac{π}{2}$

$= cos 4 (\frac{π}{2}) + sin (4 \times \frac{π}{2})$

$= cos 2 π + sin 2 π$

$= 1 + 0$

$= 1$