Question

If cosx+cos squarex =1 ,then value of sin squarex+sin raised to 4is

Answer 1

Answer:

${ \huge{ \underline{ \bf{ \red{Given}}}}}$

${ \sf{cos \: x + {cos}^{2} x = 1}}$

Find:-

${ \sf{ {sin}^{2} x+ {sin}^{4} x = </u></strong><strong><u>?</u></strong><strong><u>?</u></strong><strong><u>}}$

Solution:-

${ \to{ \sf{cos \: x + {cos}^{2}x = 1 }}}$

${ \to{ \sf{cos \: x = 1 - {cos}^{2}x }}}$

From trigonometry identity....

${ \boxed{ \sf{ {sin}^{2}x + {cos}^{2} x = 1 }}}$

${ \boxed{ \sf{ {sin}^{2} x = 1 - {cos}^{2}x }}}$

${ \to{ \sf{cos \: x = {sin}^{2}x }}}$

Now, do squaring on both sides....

${ \to{ \sf{ {(cos)}^{2} x = {( {sin}^{2}) }^{2} x}}}$

${ \to{ \sf{ {cos}^{2} x = {sin}^{4} x}}}$

${ \to{ \sf{ {sin}^{4}x - {cos}^{2}x = 0 }}}$

${ \sf{ \to{now \: keep \: 1 \: at \: both \: sides}}}$

${ \to{ \sf{ {sin}^{4} x + 1 - {cos}^{2}x = 1 }}}$

From trigonometry identity.....

${ \boxed{ \sf{ {sin}^{2}x + {cos}^{2}x = 1 }}}$

${ \to{ \sf{ {sin}^{4} x + {sin}^{2}x = 1 }}}$

${ \therefore{ \sf{ {sin}^{2}x + {sin}^{4}x = 1 }}}$

Related trigonometry identities:-

sin²θ+cos²θ=1

sec²θ- tan²θ=1

csc²θ- cot²θ=1