3(sin x-cos x)^{4}+6(sin x+cos x)^{2}+4(sin ^ | vedclass

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

Question

(B) Let $f(x) = 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$.First,simplify each term:$3(\sin x-\cos x)^{4} = 3(1-2\sin x\cos

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Let $f(x) = 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$.First,simplify each term:$3(\sin x-\cos x)^{4} = 3(1-2\sin x\cos x)^{2} = 3(1+4\sin^{2}x\cos^{2}x-4\sin x\cos x) = 3+12\sin^{2}x\cos^{2}x-12\sin x\cos x$.$6(\sin x+\cos x)^{2} = 6(1+2\sin x\cos x) = 6+12\sin x\cos x$.$4(\sin^{6}x+\cos^{6}x) = 4(\sin^{2}x+\cos^{2}x)(\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x) = 4(1)((\sin^{2}x+\cos^{2}x)^{2}-3\sin^{2}x\cos^{2}x) = 4(1-3\sin^{2}x\cos^{2}x) = 4-12\sin^{2}x\cos^{2}x$.Summing these up:$f(x) = (3+12\sin^{2}x\cos^{2}x-12\sin x\cos x) + (6+12\sin x\cos x) + (4-12\sin^{2}x\cos^{2}x)$.$f(x) = 3+6+4 + (12\sin^{2}x\cos^{2}x-12\sin^{2}x\cos^{2}x) + (-12\sin x\cos x+12\sin x\cos x) = 13$.

$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$ is equal to

Similar Questions

$1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ = $

If $\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ} = m$ and $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ} = n$,then find the value of $\frac{n}{m}$.

$\sin ^2 5^{\circ}+\sin ^2 10^{\circ}+\sin ^2 15^{\circ}+\ldots+\sin ^2 90^{\circ}$ is equal to

If one solution of the equation $\cosh x - \frac{4}{5} \sinh x = 1$ is $x = 0$,then the other solution is $x =$

If $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\sin(\alpha-\beta)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module