If A(n) = sin^n alpha + cos^n alpha,then A(1) | vedclass

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

Question

(B) Given $A(n) = \sin^n \alpha + \cos^n \alpha$. We evaluate $A(1) A(4) + A(2) A(5)$: $A(1) A(4) + A(2) A(5) = (\sin \alpha + \cos \alpha)(\sin^4 \al

Vedclass · Accepted Answer

$A(1) A(6) + A(2) A(3)$

Vedclass · Answer

(B) Given $A(n) = \sin^n \alpha + \cos^n \alpha$. We evaluate $A(1) A(4) + A(2) A(5)$: $A(1) A(4) + A(2) A(5) = (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + (\sin^2 \alpha + \cos^2 \alpha)(\sin^5 \alpha + \cos^5 \alpha)$. Since $\sin^2 \alpha + \cos^2 \alpha = 1$,this becomes: $(\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + (\sin^5 \alpha + \cos^5 \alpha)$. Now consider $A(1) A(6) + A(2) A(3)$: $A(1) A(6) + A(2) A(3) = (\sin \alpha + \cos \alpha)(\sin^6 \alpha + \cos^6 \alpha) + (\sin^2 \alpha + \cos^2 \alpha)(\sin^3 \alpha + \cos^3 \alpha)$. Using $\sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha) = \sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha$: $= (\sin \alpha + \cos \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha) + (\sin^3 \alpha + \cos^3 \alpha)$. $= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) - \sin^2 \alpha \cos^2 \alpha(\sin \alpha + \cos \alpha) + \sin^3 \alpha + \cos^3 \alpha$. $= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) - \sin^3 \alpha \cos^2 \alpha - \sin^2 \alpha \cos^3 \alpha + \sin^3 \alpha + \cos^3 \alpha$. $= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + \sin^3 \alpha(1 - \cos^2 \alpha) + \cos^3 \alpha(1 - \sin^2 \alpha)$. $= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + \sin^5 \alpha + \cos^5 \alpha$. This is exactly $A(1) A(4) + A(2) A(5)$. Thus,$A(1) A(4) + A(2) A(5) = A(1) A(6) + A(2) A(3)$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

If $A(n) = \sin^n \alpha + \cos^n \alpha$,then $A(1) A(4) + A(2) A(5) =$

Similar Questions

If $\tan 15^{\circ}+\frac{1}{\tan 75^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2a$,then the value of $\left(a+\frac{1}{a}\right)$ is:

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 $\frac{n}{m}$.

$\frac{1+\cos \theta-\sin \theta}{1+\cos \theta+\sin \theta}+\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=$

If $\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}$,then $\theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module