The value of cos(18^{circ}-A) cdot cos(18^{ci | vedclass

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

Question

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have: $\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$ $\cos

Vedclass · Accepted Answer

$\sin 54^{\circ}$

Vedclass · Answer

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have: $\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$ $\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$ Subtracting these: $(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$ Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes $\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 	imes 18^{\circ}) = \cos 36^{\circ}$. Alternatively,using $\cos(x-y)\cos(x+y) = \frac{1}{2}(\cos 2x + \cos 2y)$: $\frac{1}{2}(\cos 36^{\circ} + \cos 2A) - \frac{1}{2}(\cos 144^{\circ} + \cos 2A) = \frac{1}{2}(\cos 36^{\circ} - \cos 144^{\circ})$ Since $\cos 144^{\circ} = -\cos 36^{\circ}$,we get $\frac{1}{2}(\cos 36^{\circ} + \cos 36^{\circ}) = \cos 36^{\circ} = \sin 54^{\circ}$.

The value of $\cos(18^{\circ}-A) \cdot \cos(18^{\circ}+A) - \cos(72^{\circ}-A) \cdot \cos(72^{\circ}+A)$ is:

Similar Questions

$\sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}\right)$ is equal to

If $\cosh x = \operatorname{cosec} \theta$,then $\coth^2 \frac{x}{2} = $

If $\alpha = \frac{\sin^3 x}{\cos^2 x}$,$\beta = \frac{\cos^3 x}{\sin^2 x}$ and $\sin x + \cos x = k$,then $\alpha \sin x + \beta \cos x + 3 = $

If $\theta$ is eliminated from the equations $x = a \cos(\theta - \alpha)$ and $y = b \cos(\theta - \beta)$,then $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta)$ is equal to

The set $\{x \in R: |\cos x| \geq \sin x\} \cap \left[0, \frac{3 \pi}{2}\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module