let $$f($$θ$$)=sin2$$⁡θ$$cos2$$⁡θ$$1+4sin$$⁡$$4$$θ$$sin2$$⁡θ$$1+cos2$$⁡θ$$4sin$$⁡$$4$$θ$$1+sin2$$⁡θ$$cos2$$⁡θ$$4sin$$⁡$$4$$θ then f is

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let $$f($$θ$$)=sin2$$⁡θ$$cos2$$⁡θ$$1+4sin$$⁡$$4$$θ$$sin2$$⁡θ$$1+cos2$$⁡θ$$4sin$$⁡$$4$$θ$$1+sin2$$⁡θ$$cos2$$⁡θ$$4sin$$⁡$$4$$θ then f is

Infinity Learn · Accepted Answer

$$f($$θ$$)=$$−$$101$$−$$1101+sin2$$⁡θ$$cos2$$⁡θ$$4sin$$⁡$$2$$θ                                              $$(R1$$→$$R1-R3$$   $$R2$$ →$$R2-R3)=001$$−$$1101+sin2$$⁡θ$$cos2$$⁡θ$$4sin$$⁡$$4$$θ$$+4sin$$⁡$$4$$θ   $$(C1$$ →$$C1$$ $$+$$ $$C3)=$$−$$cos2$$⁡θ$$+1+sin2$$⁡θ$$+4sin$$⁡$$4$$θ⇒$$-2(1+2sin$$ $$4$$ θ$$)which$$ is periodic function with period $$2$$$$π$$$$4=$$π$$2$$.

let $f (θ) = |\begin{array}{ccc} \sin^{2} θ & \cos^{2} θ & 1 + 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \end{array}|$ then f is

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let f(θ)=sin2⁡θcos2⁡θ1+4sin⁡4θsin2⁡θ1+cos2⁡θ4sin⁡4θ1+sin2⁡θcos2⁡θ4sin⁡4θ then f is

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let $f (θ) = |\begin{array}{ccc} \sin^{2} θ & \cos^{2} θ & 1 + 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \end{array}|$ then f is