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[Solved] Trigonometric functions equations

Trigonometry 1 & 2

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June 15, 2021 2:13 pm

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June 15, 2021 2:13 pm

Topic starter

Staff

(@armor)

Admin

Honorable Member

317 Posts
156 153 0

Since $\small f(sinx)=sin(4n+1)x$ and $\small cosx=sin(\frac{\Pi }{2}-x),$ then $\small f(cosx)=f[sin(\frac{\Pi }{2}-x)]=sin[(4n+1)(\frac{\Pi }{2}-x)]$

$\small =sin[2nx+\frac{\Pi }{2}-(4n+1)x]=sin[\frac{\Pi }{2}-(4n+1)x]=cos(4n+1)x.$

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