Question

if sintheta +costheta =x then show that sin ^8theta +cos^8theta =4-3(x^2-1)^2/4

Answer 1

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Answer 2

Answer:

Given:

$sin\,\theta+cos\,\theta=x$

To Show: $sin^6\,\theta+cos^6\,\theta=\frac{4-3(x^2-1)^}{4}$

Consider,

$sin\,\theta+cos\,\theta=x$

Square both sides.

$(sin\,\theta+cos\,\theta)^2=x^2$

using identity , ( a + b )² = a² + b² + 2ab  

$sin^2\,\theta+cos^2\,\theta+2\,sin\,\theta\,cos\,\theta=x^2$

$1+2\,sin\,\theta\,cos\,\theta=x^2$

$2\,sin\,\theta\,cos\,\theta=x^2-1$

$sin\,\theta\,cos\,\theta=\frac{x^2-1}{2}$ ...................(1)

Now Consider,

LHS

$=sin^6\,\theta+cos^6\,\theta$

$=(sin^2\,\theta)^3+(cos^2\,\theta)^3$

using identity, a³ + b³ = ( a + b )( a²  + b² - ab )

$=(sin^2\,\theta+cos^2\,\theta)((sin^2\,\theta)^2+(cos^2\,\theta)^2-sin^2\,\theta\:cos^2\,\theta)$

$=(1)((sin^2\,\theta)^2+(cos^2\,\theta)^2-sin^2\,\theta\:cos^2\,\theta)$

$=(sin^2\,\theta)^2+(cos^2\,\theta)^2-sin^2\,\theta\:cos^2\,\theta$

using, a² + b² = ( a + b )² - 2ab

$=(sin^2\,\theta+cos^2\,\theta)^2-2\:sin^2\,\theta\:cos^2\,\theta-sin^2\,\theta\:cos^2\,\theta$

$=(1)^2-3\:sin^2\,\theta\:cos^2\,\theta$

$=1-3\:(sin\,\theta\:cos\,\theta)^2$

$=1-3\:(\frac{x^2-1}{2})^2$  ( from (1) )

$=1-\frac{3(x^2-1)^2}{2^2}$

$=\frac{4-3(x^2-1)^2}{4}$

$=RHS$

Hence proved.