Question

Solve: cosine (x + pi) = one-half over the interval left-bracket startfraction pi over 2 endfraction, pi right-bracket.

Answer 1

The values that represent the solution to $cos (\frac{\pi }{4} - x) = \frac{\sqrt{2} }{2} sin x$ ,

x ∈ [0, 2·π) are:

              $x = (\frac{\pi }{2} , \frac{3\pi }{2} )$

Sine and Cosine are trigonometric functions of an angle. The sine and cosine of an acute angle are defined in the context of a right triangle: for the specified angle, its sine is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse), and the cosine is the ratio of the length of the adjacent leg to that of the hypotenuse. For an angle θ the sine and cosine functions are denoted simply as sinθ and cos θ.

Once such a triangle is chosen, the sine of the angle is equal to the length of the opposite side, divided by the length of the hypotenuse:

Sin (α) = $\frac{opposite}{Hypotenuse}$  and  Cos (α) = $\frac{Adjacent}{Hypotenues}$

Given in the question:

The given function is presented as follows;

$cos (\frac{\pi }{4} - x) = \frac{\sqrt{2} }{2} sin x$

Where;

x ∈ [0, 2·π)

We have;

cos(A - B) = cos(A)·cos(B) - sin(A)·sin(B)

$cos (\frac{\pi }{4} ) = \frac{\sqrt{2} }{2} , sin( \frac{\pi }{4)} =\frac{\sqrt{2} }{2}$

Which gives

$cos(\frac{\pi }{4} - x ) = \frac{\sqrt{2} }{2} cos x +\frac{\sqrt{2} }{2} sinx$

Therefore, we get;

$\frac{\sqrt{2} }{2} cos x = \frac{\sqrt{2} }{2} sinx - \frac{\sqrt{2} }{2} = 0$

cos x = 0

$cos (\frac{\pi }{2} ) =0$

Therefore;

The possible solutions (x-values) to cos x = 0 are;

                             $0\leq \frac{n.\pi }{2} \leq 2\pi$

Where;

n = 1, 3, 5, ...

x ∈ [0, 2·π)

We get;   $x = \frac{\pi }{2} or \frac{3\pi }{2}$

Learn more about Cosine:

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