If sin theta + cos theta = p and sin^3 theta | vedclass

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

Question

(D) Given,$\sin 	heta + \cos 	heta = p$ $(i)$ and $\sin^3 	heta + \cos^3 	heta = q$ (ii). Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Vedclass · Accepted Answer

$-2q$

Vedclass · Answer

(D) Given,$\sin 	heta + \cos 	heta = p$ $(i)$ and $\sin^3 	heta + \cos^3 	heta = q$ (ii). Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we have: $(\sin 	heta + \cos 	heta)(\sin^2 	heta - \sin 	heta \cos 	heta + \cos^2 	heta) = q$. Since $\sin^2 	heta + \cos^2 	heta = 1$,this becomes $p(1 - \sin 	heta \cos 	heta) = q$. Thus,$1 - \sin 	heta \cos 	heta = \frac{q}{p}$,which implies $\sin 	heta \cos 	heta = 1 - \frac{q}{p}$ (iii). Squaring equation $(i)$: $(\sin 	heta + \cos 	heta)^2 = p^2$. $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = p^2$. $1 + 2 \sin 	heta \cos 	heta = p^2$. Substituting (iii) into this: $1 + 2(1 - \frac{q}{p}) = p^2$. $1 + 2 - \frac{2q}{p} = p^2$. $3 - \frac{2q}{p} = p^2$. Multiply by $p$: $3p - 2q = p^3$. Rearranging gives $p^3 - 3p = -2q$,or $p(p^2 - 3) = -2q$.

If $\sin \theta + \cos \theta = p$ and $\sin^3 \theta + \cos^3 \theta = q$,then $p(p^2 - 3)$ is equal to

Similar Questions

$\sin 20^{\circ}(4+\sec 20^{\circ})=$

If $\tan \theta + \cot \theta = 2$,then $\sin \theta$ is equal to

If $3 \sin^{2} x - 8 \sin x + 4 = 0$ and $x \in \left(\frac{\pi}{2}, \pi\right)$,then $\tan x = $

Evaluate the expression: $\sum \frac{1}{1 + x^{a-b} + x^{a-c}}$

Let $P = \{ \theta : \sin \theta - \cos \theta = \sqrt{2} \cos \theta \}$ and $Q = \{ \theta : \sin \theta + \cos \theta = \sqrt{2} \sin \theta \}$ be two sets. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module