Show without the use of a calculator, that: (180+x) (x−360^) 38^(360+x) 240^ 225^ 412^

Question

ScanSolve AI Tutor · Accepted Answer

let's simplify this trigonometric expression without a calculator.

7.3
We need to simplify the expression:

\frac{\sin(180^\circ + x) \tan(x - 360^\circ) \cos38^\circ}{\tan(360^\circ + x) \cos240^\circ \tan225^\circ \sin412^\circ}

Simplify each trigonometric term in the numerator.
- $\sin(180^\circ + x) = -\sin x$ (Sine is negative in the third quadrant)
- $\tan(x - 360^\circ) = \tan x$ (Tangent has a period of $180^\circ$ or $360^\circ$ )
- $\cos38^\circ$ remains as is for now.
Simplify each trigonometric term in the denominator.
- $\tan(360^\circ + x) = \tan x$ (Tangent has a period of $360^\circ$ )
- $\cos240^\circ = \cos(180^\circ + 60^\circ) = -\cos60^\circ = -\frac{1}{2}$ (Cosine is negative in the third quadrant)
- $\tan225^\circ = \tan(180^\circ + 45^\circ) = \tan45^\circ = 1$ (Tangent is positive in the third quadrant)
- $\sin412^\circ = \sin(360^\circ + 52^\circ) = \sin52^\circ$ (Sine has a period of $360^\circ$ )
Substitute the simplified terms back into the original expression.
$\frac{(-\sin x)(\tan x)(\cos38^\circ)}{(\tan x)(-\frac{1}{2})(1)(\sin52^\circ)}$
Cancel common terms and simplify the expression.
Cancel $\tan x$ from the numerator and denominator (assuming $\tan x \ne 0$ ):
$\frac{-\sin x \cos38^\circ}{-\frac{1}{2} \sin52^\circ}$
The negative signs cancel out:
$\frac{\sin x \cos38^\circ}{\frac{1}{2} \sin52^\circ}$
Use the co-function identity $\sin(90^\circ - \theta) = \cos \theta$ .
We know that $\sin52^\circ = \sin(90^\circ - 38^\circ) = \cos38^\circ$ .
Substitute this into the expression:
$\frac{\sin x \cos38^\circ}{\frac{1}{2} \cos38^\circ}$
Cancel $\cos38^\circ$ from the numerator and denominator (assuming $\cos38^\circ \ne 0$ ):
$\frac{\sin x}{\frac{1}{2}}$ $2 \sin x$

Show without the use of a calculator, that: (180+x) (x-360^) 38^(360+x) 240^ 225^ 412^