Differentiation (Part-38) | S N De

 $\,4(v)\,$ Prove that $\,\,\sin^254^{\circ}+\cos^272^{\circ}=\frac 34$

Sol. $\,\,\sin^254^{\circ}+\cos^272^{\circ}\\=\left(\frac{\sqrt5+1}{4}\right)^2+\left(\frac{\sqrt5-1}{4}\right)^2\\=\frac{2[(\sqrt5)^2+1^2]}{16}\\=\frac{2 \times 6}{16}\\=\frac 34$

$\,4(vi)\,$ Prove that $\,\,\cos^248^{\circ}-\sin^212^{\circ}=\frac 18(\sqrt5+1)$

Sol.  $\,\,\cos^248^{\circ}-\sin^212^{\circ}\\=\cos(48^{\circ}+12^{\circ})\cos(48^{\circ}-12^{\circ})\\=\cos60^{\circ}\cos36^{\circ}\\=\frac 12. \frac{\sqrt5+1}{4}\\=\frac 18(\sqrt5+1)$

$\,4(vii)\,$ Prove that $\,\,\cos\left(157\frac 12\right)^{\circ}=-\frac12\sqrt{2+\sqrt2}$

Sol.   $\,\,\cos\left(157\frac 12\right)^{\circ}\\=\cos\left(180^{\circ}-22\frac 12^{\circ}\right)\\=-\cos22\frac 12^{\circ}\\=-\sqrt{\cos^222\frac 12^{\circ}}\\=-\sqrt{\frac{1+\cos45^{\circ}}{2}}\\=-\sqrt{\frac{1+\frac{1}{\sqrt2}}{2}}\\=-\sqrt{\frac{2(1+\frac{1}{\sqrt2})}{4}}\\=-\frac 12\sqrt{2+\sqrt2}\,\,\,\,\,\,\,\,\text{(proved)}$

$\,4(viii)\,$ Prove that $\,\, \cos6^{\circ} \cos42^{\circ}\cos66^{\circ}\cos78^{\circ}=\frac{1}{16}$

Sol. $\,\,\cos6^{\circ} \cos42^{\circ}\cos66^{\circ}\cos78^{\circ}\\=\frac 14(2\cos6^{\circ} \cos66^{\circ} )(2\cos42^{\circ} \cos78^{\circ} )\\=\frac 14\left[\frac{\sqrt5-1}{4}+\frac 12\right] \left[-\frac 12+\frac{\sqrt5+1}{4}\right]\\=\frac 14\times \frac{\sqrt5+1}{4} \times \frac{\sqrt5-1}{4}\\=\frac{5-1}{4 \times 16}\\=\frac{1}{16}\,\,\,\text{(proved)}$

$\,4(ix)\,$ Prove that $\,\, \sin6^{\circ} \sin 42^{\circ}\sin66^{\circ}\sin78^{\circ}=\frac{1}{16}$

Sol. $\,\, \sin6^{\circ} \sin 42^{\circ}\sin66^{\circ}\sin78^{\circ}\\=\frac 14(2\sin6^{\circ}\sin66^{\circ})(2\sin 42^{\circ}\sin78^{\circ})\\=\frac 14[\cos(66^{\circ}-6^{\circ}) \cos(66^{\circ}+6^{\circ})] \\ \times [\cos(78^{\circ}-42^{\circ})-\cos(78^{\circ}+42^{\circ})]\\=\frac 14[\cos60^{\circ}-\cos72^{\circ}][\cos36^{\circ}-\cos120^{\circ}]\\=\frac 14\left[\frac 12-\frac{\sqrt5-1}{4}\right]\left[\frac{\sqrt5+1}{4}-\left(-\frac 12\right)\right]\\=\frac 14.\frac{3-\sqrt5}{4}.\frac{\sqrt5+3}{4}\\=\frac 14\times \frac{3^2-(\sqrt5)^2}{16}\\=\frac 14\times \frac{9-5}{16}\\=\frac{1}{16}\,\,\,\text{(proved)}$

$\,4(x)\,$ Prove that $\,\, 2\cos(11\frac 14^{\circ})=\sqrt{2+\sqrt{2+\sqrt{2}}}$

Sol. $\,\,11\frac 14^{\circ}=\frac 12 \times 22\frac 12^{\circ}=\frac 12 \times \frac 12\times 45^{\circ}$

Now, $\,\,2\cos11\frac 14^{\circ}\\=\sqrt{4\cos^211\frac 14^{\circ}}\\=\sqrt{2.2\cos^211\frac 14^{\circ}}\\=\sqrt{2(1+\cos22\frac 12^{\circ})}\\=\sqrt{2+2\cos22\frac 12^{\circ}}\\=\sqrt{2+\sqrt{4\cos^222\frac 12^{\circ}}}\\=\sqrt{2+\sqrt{2.2\cos^222\frac 12^{\circ}}}\\=\sqrt{2+\sqrt{2(1+\cos45^{\circ})}}\\=\sqrt{2+\sqrt{2(1+\frac{1}{\sqrt2})}}\\=\sqrt{2+\sqrt{2+\sqrt2}}\,\,\,\text{(proved)}$

$\,4(xi)\,$ Prove that $\,\, 2\sin(11\frac 14^{\circ})=\sqrt{2-\sqrt{2+\sqrt{2}}}$

Sol. $\,\,11\frac 14^{\circ}=\frac 12 \times 22\frac 12^{\circ}=\frac 12 \times \frac 12\times 45^{\circ}$

Now, $\,\,2\sin11\frac 14^{\circ}\\=\sqrt{4\sin^211\frac 14^{\circ}}\\=\sqrt{2.2\sin^211\frac 14^{\circ}}\\=\sqrt{2(1-\cos22\frac 12^{\circ})}\\=\sqrt{2-2\cos22\frac 12^{\circ}}\\=\sqrt{2-\sqrt{4\cos^222\frac 12^{\circ}}}\\=\sqrt{2-\sqrt{2.2\cos^222\frac 12^{\circ}}}\\=\sqrt{2-\sqrt{2(1+\cos45^{\circ})}}\\=\sqrt{2-\sqrt{2(1+\frac{1}{\sqrt2})}}\\=\sqrt{2-\sqrt{2+\sqrt2}}\,\,\,\text{(proved)}$

$\,4(xii)\,$ Prove that $\,\, 8\cos^27^{\circ}30'=4+\sqrt6+\sqrt2$

Sol. $\,\,8\cos^27^{\circ}30'\\=8\cos^27\frac 12^{\circ}\\=4.2\cos^27\frac 12^{\circ}\\=4(1+\cos15^{\circ})\\=4+4\cos15^{\circ}\\=4+2\sqrt{4\cos^215^{\circ}}\\=4+2\sqrt{2.2\cos^215^{\circ}}\\=4+2\sqrt{2(1+\cos30^{\circ})}\\=4+2\sqrt{2\left(1+\frac{\sqrt3}{2}\right)}\\=4+2\sqrt{2+\sqrt3}\\=4+\sqrt{8+4\sqrt3}\\=4+\sqrt{(\sqrt6)^2+(\sqrt2)^2+2.\sqrt6.\sqrt2}\\=4+\sqrt{(\sqrt6+\sqrt2)^2}\\=4+\sqrt6+\sqrt2\,\,\,\text{(proved)}$

$\,4(xiii)\,$ Prove that $\,\, 8\sin^27^{\circ}30'=4-\sqrt6-\sqrt2$

Sol. $\,\,8\sin^27^{\circ}30'\\=8\sin^27\frac 12^{\circ}\\=4.2\sin^27\frac 12^{\circ}\\=4(1-\cos15^{\circ})\\=4-4\cos15^{\circ}\\=4-2\sqrt{4\cos^215^{\circ}}\\=4-2\sqrt{2.2\cos^215^{\circ}}\\=4-2\sqrt{2(1+\cos30^{\circ})}\\=4-2\sqrt{2\left(1+\frac{\sqrt3}{2}\right)}\\=4-2\sqrt{2+\sqrt3}\\=4-\sqrt{8+4\sqrt3}\\=4-\sqrt{(\sqrt6)^2+(\sqrt2)^2+2.\sqrt6.\sqrt2}\\=4-\sqrt{(\sqrt6+\sqrt2)^2}\\=4-\sqrt6-\sqrt2\,\,\,\text{(proved)}$

$\,4(xiv)\,$ Prove that $\,\, \sin x=2^n.\cos\frac x2.\cos\frac{x}{2^2}.\cos\frac{x}{2^3}\cdots \cos\frac{x}{2^n}\sin\frac{x}{2^n}$

Sol. $\,\,2^n.\cos\frac x2.\cos\frac{x}{2^2}.\cos\frac{x}{2^3}\cdots \cos\frac{x}{2^n}\sin\frac{x}{2^n}\\=2^{n-1}\cos\frac x2.\cos\frac{x}{2^2}.\cos\frac{x}{2^3}\cdots \cos\frac{x}{2^{n-1}}\\ \times\left(2\cos\frac{x}{2^n}\sin\frac{x}{2^n}\right)\\=2^{n-1}\cos\frac x2.\cos\frac{x}{2^2}.\cos\frac{x}{2^3}\cdots \cos\frac{x}{2^{n-1}}\sin\frac{x}{2^{n-1}}\\=2^{n-2}\cos\frac x2.\cos\frac{x}{2^2}.\cos\frac{x}{2^3}\cdots \cos\frac{x}{2^{n-2}}\\ \times \left(2\cos\frac{x}{2^{n-1}}\sin\frac{x}{2^{n-1}}\right)\\=2^{n-2}\cos\frac x2.\cos\frac{x}{2^2}.\cos\frac{x}{2^3}\cdots \cos\frac{x}{2^{n-2}}\sin\frac{x}{2^{n-2}}\\=\cdots \cdots \cdots \\=2\cos\frac x2(2\cos\frac{x}{2^2}\sin\frac{x}{2^2})\\=2\cos\frac x2\sin\frac x2\\=\sin x\,\,\,\text{(proved)}$

$\,4(xv)\,$ Prove that, $\,\,\sin^6\frac{\theta}{2} +\cos^6\frac{\theta}{2}=\frac 14(1+3\cos^2\theta)$

Sol. $\,\,\sin^6\frac{\theta}{2} +\cos^6\frac{\theta}{2}\\=(\sin^2\frac{\theta}{2})^3+(\cos^2\frac{\theta}{2})^3\\=(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2})(\sin^4\frac{\theta}{2}+\cos^4\frac{\theta}{2}\\-\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2})\\=1.\left[(\sin^2\frac{\theta}{2})^2+(\cos^2\frac{\theta}{2})^2-\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}\right]\\=(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2})^2-2\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}\\-\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}\\=1^2-3\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}\\=1-\frac 34(2\sin\frac{\theta}{2}\cos\frac{\theta}{2})^2\\=1-\frac 34\sin^2\theta\\=1-\frac 34(1-\cos^2\theta)\\=1-\frac 34+\frac 34\cos^2\theta\\=\frac 14+\frac 34\cos^2\theta\\=\frac 14(1+3\cos^2\theta)\quad\text{(proved)}$