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Which values represent solutions to cosine (startfraction pi over 4 endfraction minus x) = startfraction startroot 2 endroot over 2 endfraction sine x, where x is an element of [0, 2pi)?

Sagot :

The given trigonometric function consists of trigonometric values of common angles.

The values that represent the solution to [tex]cos(\frac{\pi }{4} -x)=\frac{\sqrt{2} }{2} .sinx[/tex], x ∈ [0, 2·π) are; [tex]x=({\frac{\pi }{2} or \frac{3.\pi }{2} )[/tex].

What are trigonometric functions?

  • Trigonometric functions are real functions in mathematics that connect an angle of a right-angled triangle to ratios of two side lengths.
  • They are widely utilized in all geosciences, including navigation, solid mechanics, celestial mechanics, geodesy, and many more.

The given function is:  [tex]cos(\frac{\pi }{4} -x)=\frac{\sqrt{2} }{2} .sinx[/tex]

Where; x ∈ [0, 2·π).

We have; cos(A - B) = cos(A)·cos(B) - sin(A)·sin(B).

  • [tex]cos(\frac{\pi }{4} )=\frac{\sqrt{2} }{2}, sin(\frac{\pi }{4} )=\frac{\sqrt{2} }{2}[/tex]

Which gives:

  • [tex]cos(\frac{\pi }{4}-x )=\frac{\sqrt{2} }{2}. cosx+\frac{\sqrt{2} }{2}.sinx[/tex]

So, we obtain:
[tex]\frac{\sqrt{2} }{2} .cosx=\frac{\sqrt{2} }{2}.sinx-\frac{\sqrt{2} }{2} .sinx=0\\cosx=0\\cos\frac{\pi }{2} =0[/tex]

Therefore, the possible solutions (x-values) to cos x = 0 are:

[tex]0[/tex] ≤ [tex]\frac{n.\pi }{2}[/tex] ≤ [tex]2.\pi[/tex]

Where, n = 1, 3, 5, .., x ∈ [0, 2·π)

We get:  [tex]x=({\frac{\pi }{2} or \frac{3.\pi }{2} )[/tex].

Know more about trignometric funcions here:

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