3left[ sin^4left( frac{3pi}{2} - alpha right) | vedclass

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(B) આપેલ પદાવલિ: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \

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(B) આપેલ પદાવલિ: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right]$ત્રિકોણમિતીય રિડક્શન સૂત્રોનો ઉપયોગ કરતા:$\sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha$,$\sin(3\pi + \alpha) = -\sin \alpha$,$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$,$\sin(5\pi - \alpha) = \sin \alpha$આ કિંમતો મૂકતા:$= 3\left[ (-\cos \alpha)^4 + (-\sin \alpha)^4 \right] - 2\left[ (\cos \alpha)^6 + (\sin \alpha)^6 \right]$$= 3(\cos^4 \alpha + \sin^4 \alpha) - 2(\cos^6 \alpha + \sin^6 \alpha)$નિત્યસમ $\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$ અને $\sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha$ નો ઉપયોગ કરતા:$= 3(1 - 2\sin^2 \alpha \cos^2 \alpha) - 2(1 - 3\sin^2 \alpha \cos^2 \alpha)$$= 3 - 6\sin^2 \alpha \cos^2 \alpha - 2 + 6\sin^2 \alpha \cos^2 \alpha$$= 3 - 2 = 1$

$3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right] = $

Similar Questions

$0 < \theta < \frac{\pi}{2}$ માટે,$\sum_{m=1}^6 \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) = 4 \sqrt{2}$ ના ઉકેલ(ઓ) છે:

$\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ} = $

ધારો કે $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ જ્યાં $k = 1, 2, 3, ...$ છે. તો તમામ $x \in R$ માટે,$f_4(x) - f_6(x)$ ની કિંમત કેટલી થાય?

જો $0 < \theta < \frac{\pi}{2}$ અને $\sin \theta \cos \theta = \frac{12}{25}$ હોય,તો $\sin^4 \theta + \cos^4 \theta$ ની કિંમત શોધો.

જો $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$ હોય,તો $\sinh \alpha=$

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