cos^2 48^{circ} - sin^2 12^{circ} ની કિંમત શો | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) નિત્યસમ $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$ નો ઉપયોગ કરતા:$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{

Vedclass · Accepted Answer

$\frac{\sqrt{5}+1}{8}$

Vedclass · Answer

(D) નિત્યસમ $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$ નો ઉપયોગ કરતા:$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{\circ} - 12^{\circ})$$= \cos 60^{\circ} \cos 36^{\circ}$$\cos 60^{\circ} = \frac{1}{2}$ અને $\cos 36^{\circ} = 1 - 2\sin^2 18^{\circ}$ હોવાથી:$= \frac{1}{2} (1 - 2\sin^2 18^{\circ})$$= \frac{1}{2} \left( 1 - 2 \left( \frac{\sqrt{5}-1}{4} \right)^2 \right)$$= \frac{1}{2} \left( 1 - 2 \left( \frac{5 + 1 - 2\sqrt{5}}{16} \right) \right)$$= \frac{1}{2} \left( 1 - \frac{6 - 2\sqrt{5}}{8} \right)$$= \frac{1}{2} \left( \frac{8 - 6 + 2\sqrt{5}}{8} \right)$$= \frac{1}{2} \left( \frac{2 + 2\sqrt{5}}{8} \right) = \frac{1 + \sqrt{5}}{8}$

$\cos^2 48^{\circ} - \sin^2 12^{\circ}$ ની કિંમત શોધો,જો $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ હોય.

Similar Questions

$\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ નું મૂલ્ય શોધો.

જો $\cos (\theta+\phi)=\frac{3}{5}$ અને $\sin (\theta-\phi)=\frac{5}{13}$,જ્યાં $0 < \theta, \phi < \frac{\pi}{4}$,તો $\cot (2 \theta)$ ની કિંમત શોધો:

જો $\alpha, \beta, \gamma$ કોઈ પણ ત્રણ ખૂણાઓ હોય,તો $\cos \alpha + \cos \beta - \cos \gamma - \cos (\alpha + \beta + \gamma) =$

જો $0 < A < B < \frac{\pi}{4}$,$\cos (A+B) = \frac{11}{61}$ અને $\sin (A-B) = \frac{24}{25}$ હોય,તો $\sin 2A + \sin 2B = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module