Question

Sin101-sin19/cos71 - cos11​

Answer 1

Відповідь:

-1

Пояснення:

2cos(101+19/2)sin(101-19/2)

2cos 120/2 sin 82/2

-2sin (71+11/2) sin (71-11/2)

-2sin 82/2 sin 60/2

2cos60 (90-30) * sin41 / -2sin 41 * sin 30

-sin 30 * sin 41/ sin 41 * sin 30

-sin 41 / sin 41 = -1

Answer 2

To simplify the expression $\frac{\sin(101^\circ) - \sin(19^\circ)}{\cos(71^\circ) - \cos(11^\circ)}$, you can use the trigonometric identity:

$\sin(A) - \sin(B) = 2\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right)$

So, for your expression:

$\frac{\sin(101^\circ) - \sin(19^\circ)}{\cos(71^\circ) - \cos(11^\circ)} = \frac{2\sin\left(\frac{101^\circ - 19^\circ}{2}\right)\cos\left(\frac{101^\circ + 19^\circ}{2}\right)}{2\sin\left(\frac{71^\circ - 11^\circ}{2}\right)\cos\left(\frac{71^\circ + 11^\circ}{2}\right)}$

Now, simplify further:

$\frac{\sin(41^\circ)\cos(60^\circ)}{\sin(30^\circ)\cos(41^\circ)}$

Notice that $\cos(60^\circ) = \frac{1}{2}$ and $\sin(30^\circ) = \frac{1}{2}$, so you can simplify even more:

$\frac{\sin(41^\circ)\cdot\frac{1}{2}}{\frac{1}{2}\cdot\cos(41^\circ)}$

Now, cancel out the common factors of $\frac{1}{2}$:

$\frac{\sin(41^\circ)}{\cos(41^\circ)}$

This is equal to $\tan(41^\circ)$.

So, $\frac{\sin(101^\circ) - \sin(19^\circ)}{\cos(71^\circ) - \cos(11^\circ)} = \tan(41^\circ)$.