How do you find the other trigonometric functions given $sin (\frac{π}{2} - x) = \frac{\sqrt{2}}{2}, sin x = - \frac{\sqrt{2}}{2}$ $sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2$ ?

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How do you find the other trigonometric functions given $sin (\frac{π}{2} - x) = \frac{\sqrt{2}}{2}, sin x = - \frac{\sqrt{2}}{2}$ $sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2$ ?

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Answer 1 · 2017-08-21T09:59:38

$see explanation$ $"see explanation"$

Explanation:

$given sin (\frac{π}{2} - x) = \frac{\sqrt{2}}{2}$ $"given "sin(pi/2-x)=sqrt2/2$

$\Rightarrow cos x = \frac{\sqrt{2}}{2} and sin x = - \frac{\sqrt{2}}{2}$ $rArrcosx=sqrt2/2" and "sinx=-sqrt2/2$

$\Rightarrow x is in the fourth quadrant$ $rArrx" is in the fourth quadrant"$

$∙ x sec x = \frac{1}{cos x} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ $•color(white)(x)secx=1/cosx=1/(sqrt2/2)=2/sqrt2=sqrt2$

$∙ x csc x = \frac{1}{sin x} = \frac{1}{- \frac{\sqrt{2}}{2}} = - \frac{2}{\sqrt{2}} = - \sqrt{2}$ $•color(white)(x)cscx=1/sinx=1/(-sqrt2/2)=-2/sqrt2=-sqrt2$

$∙ x tan x = \frac{sin x}{cos x} = \frac{- \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = - 1$ $•color(white)(x)tanx=sinx/cosx=(-sqrt2/2)/(sqrt2/2)=-1$

$∙ x cot x = \frac{1}{tan x} = - 1$ $•color(white)(x)cotx=1/tanx=-1$

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How do you find the other trigonometric functions given $sin (\frac{π}{2} - x) = \frac{\sqrt{2}}{2}, sin x = - \frac{\sqrt{2}}{2}$ $sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2$ ?

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How do you find the other trigonometric functions given sin(π2−x)=√22,sinx=−√22sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2?

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How do you find the other trigonometric functions given $sin (\frac{π}{2} - x) = \frac{\sqrt{2}}{2}, sin x = - \frac{\sqrt{2}}{2}$ $sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2$ ?