જો f(theta) = cos^3 theta + cos^3 left(frac{2 | vedclass

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(A) આપણે નિત્યસમ $\cos^3 \alpha + \cos^3 \left(\alpha + \frac{2\pi}{3}\right) + \cos^3 \left(\alpha + \frac{4\pi}{3}\right) = \frac{3}{4} \cos(3\alpha

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$\frac{3(\sqrt{5}-1)}{16}$

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(A) આપણે નિત્યસમ $\cos^3 \alpha + \cos^3 \left(\alpha + \frac{2\pi}{3}ight) + \cos^3 \left(\alpha + \frac{4\pi}{3}ight) = \frac{3}{4} \cos(3\alpha)$ નો ઉપયોગ કરીએ છીએ.આપેલ છે કે $f(	heta) = \cos^3 	heta + \cos^3 \left(	heta + \frac{2\pi}{3}ight) + \cos^3 \left(	heta - \frac{2\pi}{3}ight)$.કારણ કે $\cos \left(	heta - \frac{2\pi}{3}ight) = \cos \left(	heta + \frac{4\pi}{3}ight)$,તેથી $f(	heta) = \frac{3}{4} \cos(3	heta)$ થાય.$	heta = \frac{\pi}{5}$ મૂકતા,$f\left(\frac{\pi}{5}ight) = \frac{3}{4} \cos \left(\frac{3\pi}{5}ight)$ મળે.આપણે જાણીએ છીએ કે $\cos \left(\frac{3\pi}{5}ight) = \frac{1-\sqrt{5}}{4}$.તેથી,$f\left(\frac{\pi}{5}ight) = \frac{3}{4} \left(\frac{1-\sqrt{5}}{4}ight) = \frac{3(1-\sqrt{5})}{16}$.

જો $f(\theta) = \cos^3 \theta + \cos^3 \left(\frac{2\pi}{3} + \theta\right) + \cos^3 \left(\theta - \frac{2\pi}{3}\right)$ હોય,તો $f\left(\frac{\pi}{5}\right) = $

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જો કાટકોણ ત્રિકોણની બાજુઓ ${\cos2\alpha + \cos2\beta + 2\cos(\alpha + \beta )}$ અને ${\sin2\alpha + \sin2\beta + 2\sin(\alpha + \beta )}$ હોય,તો કર્ણની લંબાઈ શોધો:

કોઈપણ $\theta \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)$ માટે,પદાવલિ $3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4\sin^6 \theta$ ની કિંમત શોધો.

સરવાળો શોધો: $\sin^2(3^{\circ}) + \sin^2(6^{\circ}) + \sin^2(9^{\circ}) + \dots + \sin^2(84^{\circ}) + \sin^2(87^{\circ}) + \sin^2(90^{\circ})$

જો $\cos \alpha = \operatorname{sech} \beta$ હોય,તો $\beta =$

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