What is sin⁡(x)+cos⁡(x) in terms of sine?

Question

Bubich13 · Accepted Answer

Explanation:
Using Pythagorean Identity

\sin^{2} x + \cos^{2} x = 1, so \cos^{2} x = 1 - \sin^{2} x

\cos x = \pm \sqrt{1 - \sin^{2} x}

\sin x + \cos x = \sin x \pm \sqrt{1 - \sin^{2} x}

Using complement / cofunction identity

\cos x = \sin (\frac{π}{2} - x)

\sin x + \cos x = \sin x + \sin (\frac{π}{2} - 2)

lovagwb · Accepted Answer

Explanation  Suppose that sin⁡x+cos⁡x=Rsin⁡(x+α)  Then  sin⁡x+cos⁡x=Rsin⁡xcos⁡α+Rcos⁡xsin⁡α  =(Rcos⁡α)sin⁡x+(Rsin⁡α)cos⁡x  The coefficients of sin⁡x and of cos⁡x must be equal so  Rcos⁡α=1Rsin⁡α=1  Squaring and adding, we get  R2cos2α+R2sin2α=2  so  R2(cos2α+sin2α)=2  R=2  And now  cos⁡α=12  sin⁡α=12  So  α=cos−1(12)=π4  sin⁡x+cos⁡x=2sin⁡(x+π4)

Vasquez · Accepted Answer

Explanation:
Using pythagorean identity,
$\sin^{2} x + \cos^{2} x = 1$
$So, \cos^{2} x = 1 - \sin^{2} x$
By taking square root on both the sides,
$\cos x + \sin x = \sin \pm \sqrt{1 - \sin^{2} x}$
Using complement or cofunction identity,
$\cos x = \sin (\frac{π}{2} - x)$
$\sin x + \cos x = \sin x + \sin (\frac{π}{2} - x)$
Thus, the expression for $\sin x + \cos x$ in terms of sine is $\sin x + \sin (\frac{π}{2} - x) .$

What is \sin(x)+\cos(x) in terms of sine?

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