1. Tardigrade
  2. Question
  3. Mathematics
  4. If cos α + sin α =(3/4),then sin6 α + cos6 α is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $cos \alpha + sin \alpha$ =$\frac{3}{4}$,then $sin^6\, \alpha + cos^6 \, \alpha$ is equal to

KEAMKEAM 2013Trigonometric Functions

Solution:

Given, $\sin \,\alpha+\cos\, \alpha=\frac{3}{4}\,\,\,\,\,\dots(i)$
Now, $\sin ^{6}\, \alpha+\cos ^{6} \,\alpha$
$=\left(\sin ^{2} \,\alpha\right)^{3}+\left(\cos ^{2} \alpha\right)^{3}=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)$
$\left(\sin ^{4} \alpha+\cos ^{4} \alpha-\sin ^{2} \alpha \cdot \cos ^{2} \alpha\right)$
$=1 \cdot\left\{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-3 \sin ^{2} \alpha \cdot \cos ^{2} \alpha\right\}$
$\left(\because \,\sin ^{2} \alpha+\cos ^{2} \alpha=1\right) $
$=\left\{1-\frac{3}{4}(\sin 2 \alpha)^{2}\right\}\,\,\,\,\,\dots(ii)$
On squaring both sides of Eq. (i), we get
$\left(\sin ^{2} \,\alpha+\cos ^{2}\, \alpha\right)+2 \sin \alpha \cdot \cos \alpha=\frac{9}{16}$
$\Rightarrow \, \sin 2 \alpha=\frac{9}{16}-1=\frac{-7}{16}$
$ \Rightarrow (\sin 2 \alpha)^{2}=\frac{49}{256}$
On putting this value in Eq. (ii), we get
$\sin ^{6} \,\alpha+\cos ^{6} \,\alpha=\left\{1-\frac{3}{4} \times \frac{49}{756}\right\}=\left(1-\frac{147}{1024}\right)$
$=\frac{877}{1024}$