1. Simplify the 1.1. $$ \frac{\sin 135^{\circ} \cdot \cos 22}{ an 31} $$ 2. $$ \frac{\sin \left(360^{\circ}-x ight) \cdot t 2}{\cos \left(180^{\circ}+x ight) an } $$

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1. Simplify the 1.1. $$ \frac{\sin 135^{\circ} \cdot \cos 22}{	an 31} $$ 2. $$ \frac{\sin \left(360^{\circ}-xight) \cdot t 2}{\cos \left(180^{\circ}+xight) 	an } $$

UpStudy AI · Accepted Answer

1. We begin with
$$
\frac{\sin 135^\circ\cdot \cos 22^\circ}{\tan 31^\circ}\,.
$$

• First, note that
$$
\sin135^\circ=\sin(180^\circ-45^\circ)=\sin45^\circ=\frac{\sqrt{2}}{2}\,.
$$

• Also, recall that
$$
\tan31^\circ=\frac{\sin31^\circ}{\cos31^\circ}\,.
$$

• Thus we may write
$$
\frac{\sin 135^\circ\cdot \cos 22^\circ}{\tan 31^\circ}
=\frac{\frac{\sqrt{2}}{2}\cos 22^\circ}{\frac{\sin31^\circ}{\cos31^\circ}}
=\frac{\sqrt{2}}{2}\cos 22^\circ\cdot\frac{\cos31^\circ}{\sin31^\circ}\,.
$$

• Therefore, one acceptable simplified result is
$$
\frac{\sqrt{2}\cos22^\circ\cos31^\circ}{2\sin31^\circ}\,.
$$

2. In the expression
$$
\frac{\sin (360^\circ-x)\cdot \tan2x}{\cos (180^\circ+x)\tan x}\,,
$$
we assume that the intended arguments of the tangent functions are $$2x$$ in the numerator and $$x$$ in the denominator.

• First, use the identity
$$
\sin (360^\circ-x)=-\sin x\,.
$$
(To check: since $$\sin 360^\circ=0$$ and $$\cos 360^\circ=1$$, one may also write $$\sin (360^\circ-x)=0\cdot\cos x -1\cdot\sin x=-\sin x$$.)

• Next, recall that
$$
\cos (180^\circ+x)=-\cos x\,.
$$

• Substitute these into the given expression:
$$
\frac{\sin (360^\circ-x)\cdot \tan2x}{\cos (180^\circ+x)\tan x}
=\frac{(-\sin x)\cdot \tan2x}{(-\cos x)\tan x}
=\frac{\sin x\cdot \tan2x}{\cos x\cdot \tan x}\,.
$$

• But
$$
\frac{\sin x}{\cos x}=\tan x\,.
$$

• Hence we have
$$
\frac{\sin x\cdot \tan2x}{\cos x\cdot \tan x}
=\tan x\cdot \frac{\tan2x}{\tan x}
=\tan2x\,.
$$

Thus the simplified form is
$$
\tan2x\,.
$$
1. $$ \frac{\sqrt{2}\cos22^\circ\cos31^\circ}{2\sin31^\circ} $$ 2. $$ \tan2x $$

1. Simplify the 1.1. \( \frac{\sin 135^{\circ} \cdot \cos 22}{\tan 31} \) 2. \( \frac{\sin \left(360^{\circ}-x\right) \cdot t 2}{\cos \left(180^{\circ}+x\right) \tan } \)

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