Question

if sin theta +cos theta =a, then sin^4 theta + cos^ theta = pls help.... ​

Answer 1

sin © + cos © = a
Squaring both sides,
sin^2 © + cos^2 © + 2sin©cos© = a^2

We know that sin^2 © + cos^2 © = 1

So,
1 + 2sin© cos © = a^2
Therefore,

sin© cos © = a^2 - 1 / 2

Now,
Let's get back to the original
equation,

where, we got
sin^2 © + cos^2 © + 2sin© cos © = a^2
Put value of sin©cos© as found
We get,
sin^2 © + cos^2 © = 1
Lol we already knew this !

So, let's continue

Again squaring both sides,
sin^4 © + cos^4 © + 2sin^2 © cos^2 © = 1
sin^4 © + cos^4 © = 1-2sin^2 © cos^2 ©
Put the value of sin© cos © as found earlier

Sin^4 © + cos^4 © = 1-2*(a^2-1)^2/4
= 1-(a^2-1)^2/2
= 2-(a^2-1)^2/2
= 2-a^4+2a^2-1/2
= -a^4 + 2a^2 + 1 / 2

Hope this helps you !

Answer 2

Answer:

Given that, sin θ – cos θ = 0

sin θ = cos θ

sin θ / cos θ = tan θ

=> tan 45° = 1

=> θ = 45°

Now,

sin⁴ θ + cos⁴ θ

= ( sin² θ )² + ( cos² θ )²

= ( sin² 45° )² + ( cos² 45° )²

= [ ( 1 / √2 )² ]² + [ ( 1 / √2 )² ]²

= ( 1 / 2 )² + ( 1 / 2 )²

= ( 1 / 4 ) + ( 1 / 4 )

= ( 1 + 1 ) / 4

= 2 / 4

= 1 / 2