Question

sin3(21)+cos3(19)
/sin(21)+cos(19)​

Answer 1

Answer:

sin3(21)+cos3(19)/ sin(21)+cos(19)

= (sin21 + cos19)(sin2(21)+cos2(19)-sin21cos19)/sin(21)+cos(19)

= sin2(21) + cos2(19) - sin21cos19

...

Answer 2

$\dfrac{sin^{3} 21 + cos^{3} 19}{sin 21 + cos19}$ = (1 - sin 21 cos 19)

Step-by-step explanation:

$\dfrac{sin^{3} 21 + cos^{3} 19}{sin 21 + cos19}$

Now we know the formula of of $a^{3} + b^{3}$ = (a+b)($a^{2}$ -ab +$b^{2}$)

Thus putting this value in the above we get,

$\dfrac{sin^{3} 21 + cos^{3} 19}{sin 21 + cos19}$   =   $\dfrac{(sin 21 + cos19)(sin^{2} 21 - sin 21 cos 19+ cos^{2} 19)}{sin 21 + cos19}$

Crossing out sin 21+ cos 19 in numerator and denominator we get,

$(sin^{2} 21 - sin 21 cos 19+ cos^{2} 19)$

Now we know,

$(sin^{2} x+ cos^{2} x)$ =1

Therefore putting this value we get

$(sin^{2} 21 - sin 21 cos 19+ cos^{2} 19)$ = (1 - sin 21 cos 19)