The value of sin ^6(theta) + cos ^6(theta) + | vedclass

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

Question

(B) We use the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.
Then $\sin^6 	heta + \cos

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We use the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.
Then $\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3$.
$= (\sin^2 	heta + \cos^2 	heta)((\sin^2 	heta)^2 - \sin^2 	heta \cos^2 	heta + (\cos^2 	heta)^2)$.
Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:
$= 1 \cdot (\sin^4 	heta - \sin^2 	heta \cos^2 	heta + \cos^4 	heta)$.
We know that $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta$.
Substituting this back:
$\sin^6 	heta + \cos^6 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta - \sin^2 	heta \cos^2 	heta = 1 - 3 \sin^2 	heta \cos^2 	heta$.
Therefore,$\sin^6 	heta + \cos^6 	heta + 3 \sin^2 	heta \cos^2 	heta = (1 - 3 \sin^2 	heta \cos^2 	heta) + 3 \sin^2 	heta \cos^2 	heta = 1$.

The value of $\sin ^6(\theta) + \cos ^6(\theta) + 3 \sin ^2(\theta) \cos ^2(\theta)$ is

Similar Questions

If $5 \sinh x - \cosh x = 5$,then one of the values of $\tanh x$ is

In $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{(\sin^2 A + \sin^2 B)}{(\sin^2 A - \sin^2 B)} \sin(A - B) = $

If $\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1$,then the value of $\left[ \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} \right]$ is (where $[.]$ denotes the greatest integer function).

If $\tan \theta + \sec \theta = e^x$,then $\cos \theta$ equals

If $\sinh u = \tan \theta$,then $\cosh u$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module